Schwarzschild exteroior solution describes the structure of spacetime outside a matter distribution. Since there is no matter beyond the volume of the matter distribution, the energy-momentum tensor or stress-energy tensor is basically zero. So the very solution is actually a vacuum solution.

When you solve the Einstein field equation of vacuum, you will come to a solution 

e^\nu = e^{-\lambda} = 1+A / r      where A is a constant of integration.

In the weak field limit, the metric potential g_{tt} = e^\nu has to reduce to Newtonian case where   e^\nu=1+2\phi / c^2 = 1+A / r        where \phi = Newtonian poten. = - GM / r

therefore  1 + A / r  = 1 + 2\phi / c^2 =1 - 2GM / r c^2 => A = - 2GM / c^2.

Hence the final form of Schwarzschild exterior metric potentials are 

e^\nu = e^{-\lambda} = 1 - 2GM / r c^2.

What I want to suggest you is that the constant of integration A is determined is such a way that it reduces to Newtonian gravitational potential in weak field limit. This is why the mass M comes in play even in the vacuum solution. The physical meaning is that the neighboring exterior vacuum spacetime is disturb by the the mass distribution and the amount of disturbance depends on the total mass of the disbution.

the situation is the same as Laplase equation in electrostatics   when you apply the equation, the media contains no charge but if you remember we can use the eq. for a charge in which the the volume doesnt contain the charge- in S space exactly we assume the volume which doesnt contain the central mass

Far away from the mass space is flat as you have rightly expected. There is no dependence of metric on the mass. But just outside the "Schwarzchild radius" the object metric "feels" the effect of a nearby massive objects which bends space-time appropriately. You can assume a metric of the general form,

ds2= A(r) dt2 - B(r) dr2 - r2 (dθ2 + sin2θ dφ2).

Then solve for unknown functions A(r) and B(r) using Rμν=0. This simple looking equation has a lot of derivatives. A(r) and B(r) is thus determined via differential equations one of them is A'B + AB'=0, which implies AB= constant. Say AB=α, and then a second one d(rA)/dr=α. This differential equation (second one) will give another integration constant rA=α(r+k). If we keep these two (k, α) constants arbitrary then the solution is perfectly general.

But then in Physics we always construct particular solutions for a given special case. In this case we want to find out the integration constants outside a spherical mass M. For this special case, α=c2 and k=-2GM/c2. This gives desired form of Schwarzschild metric.

The problem of finding point-like sources for the Schwarzschild metric (analogue to that of the Dirac delta source for the Coulomb potential in Newtonian Gravity) is an unsolved (or unsolvable) one. See the discussion in the book of T. Ortin: "Gravity and Strings" (Section 7.2). Cambridge University Press (2004) and references cited therein. (See also section 8.4 of the same reference for the similar problem in the case of the Reissner-Nordström space time).

I agree with the answer given by those responded so far and nothing much for me to add.

I wish if I could say any thing on this answer . Perhaps some one else can explain and I am interested to know.

Adding to the response of Mojtaba Shahbazi:

The Schwarzschild solution is generally presented as a vacuum solution, analogous to the potential in empty space around a charge. The parameters in the solution (2Gm/c2) can be regarded as fixed by a boundary condition very close to the charge.

In the EM case the corresponding potential is e/r. Everywhere except at r = 0 this is a vacuum solution of the Maxwell equations. However, the fact that there is a charge source present demands that the Laplacian of the potential does not vanish everywhere. In the case that the Laplacian is expressed in spherical coordinates this is achieved by expressing the potential as e/|r|, the Laplacian of which is singular at the origin.  Generally one leaves off the modulus sign, since it is implicitly understood. (Note that formally the spherical coordinate Laplacian of e/r without the modulus sign is zero everywhere.)

In Schwarzschild's 1916 paper it seems (to me at least) that he understood his solution to the Einstein equations in the same way. Thus, although he expressed the solution in terms of the coordinate r, he implicitly meant |r|. The Einstein equations acting on a Schwarzschild metric expressed in terms of |r| gives a singularity when r = 0, corresponding to a mass (and, as it turns out, 'hoop pressure') singularity there. In the 'Schwarzschild coordinates', this corresponds to a singularity in the stress-energy tensor at the event horizon, analogous to a charge singularity at the origin when the potential is e/|r|. 

The currently accepted mainstream position however is that no such physical singularity exists at the event horizon - the stress-energy tensor is zero there - and that the coordinates can be meaningfully extended inside the black hole. Note that the same reasoning applied to a charge would remove the singularity at the location of a charge, and permit extension of the coordinates 'inside' the charge.

Then Wheeler's 'mass without mass' becomes 'charge without charge'.

I would just like to add that the solution with a point particle stress energy tensor (delta function) has been derived as well. It corresponds to the Schwarzschild solution in isotropic coordinates and features attraction at large distances and repulsion at short distances. See the following work by Katanaev for details:

https://arxiv.org/abs/1207.3481

I have asked myself that one many times,  if one chooses a scalar field as source then many unphysical properties of the schwarzshild solution such as event horizon disappear,  also the spacetime is not necessarily static,  I found the best known non-static case in 1985.....

@Zlotnikov: "I would just like to add that the solution with a point particle stress energy tensor (delta function) has been derived as well. It corresponds to the Schwarzschild solution in isotropic coordinates and features attraction at large distances and repulsion at short distances."

I have looked at the paper and I think there is a huge problem. The delta function is put at r=0, but the solution makes r the radial coordinate of the isotropic form of the Schwarzschild metric. r=0 in that form actually corresponds to infinity (the circumference of a circle with r=0 in the isotropic line element is infinite)! This then means that what the delta function describes is not a point mass (as the author claims) but a spherical shell at infinity. According to Birkhoff's theorem, we would still expect the Schwarzschild solution to arise, but clearly the solution does not quite describe what its author wants it to describe.

What would be needed is a solution, where the delta function is at (in the notatation of the paper) rho=0, where rho is the standard radial Schwarzschild coordinate (also called circumferential coordinate), i.e. at the position where the circumference of a circle becomes zero. This is not describable within the isotropic form of the metric which does not reach below the event horizon. I believe it to be difficult if not impossible to obtain such a solution, because rho=0 is the position of the true singularity, spacetime ceases to exist there, so it is problematic to imagine a mass point sitting there either (it would be strictly nowhere...).

@Kassner I see, thank you for the explanation!

If one calculates the energy-momentum tensor in the sense of distributions, it is possible to cover the Schwarzschild singularity. See Balasin and Nachbagauer, Class Quantum Grav. 10 (1993) 2271-2278.