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http://www.hamamatsu.com/resources/pdf/ssd/si_pd_circuit_e.pdf    

1 candela = 1 lumen/steradian

Now need to go from lumens to Watt.

Watts measure the amount of energy required to light products, whereas lumens measure the amount of light produced. The more lumens in a light bulb, the brighter the light. This is related to perception.

You will see sometimes

P(W) = ΦV(lm) / η(lm/W)

Where η(lm/W) is an efficiency.  however that will depend on your spectrum and details of the LED.

From Hyperphyscis.

Luminous Flux (Φv) is energy per unit time (dQ/dt) that is radiated from a source over visible wavelengths. More specifically, it is energy radiated over wavelengths sensitive to the human eye, from about 330 nm to 780 nm. Thus, luminous flux is a weighted average of theRadiant Flux in the visible spectrum. It is a weighted average because the human eye does not respond equally to all visible wavelengths.

The sensitivity of the eye peaks at 555 nm and falls off to approximately 10-4 at 380 and 750 nm. This constitutes the range of daylight sensitivity, or photopic vision. The eye's nighttime sensitivity, called scotopic vision, shifts toward the blue end of the visible, peaking at 507 nm and falling to 10-4 at 340 and 670 nm. This weighting factor, or luminous efficacy (Vλ), allows for conversion of Radiant Flux to Luminous Flux at any wavelength. In the photopic region, the peak at 555 nm is assigned a conversion value of 683 lumens per Watt. The lumen is the unit of luminous flux, and is defined in terms of the candela, an SI base unit like the meter or second. 1 lumen is defined to be 1/4π candela, the SI base unit of Luminous Intensity. 

Unfortunately for you since your wavelength of interest is not at 555 nm. So you need to know the luminous efficiency as a function of wavelength,  You can look up tables from different places.

http://hyperphysics.phy-astr.gsu.edu/hbase/vision/efficacy.html#c1

Hyperphysics again is convenient. 

So at this point you have a couple of choices. Just going from the data sheet the they give a large range.  So if you just need an estimate you can make an assumption about the spectrum and use the tables.  

the hyperphysics example is 

Since the eye does not see all wavelengths equally well, the efficacy curve is a very important way to determine the Luminous Flux from a source. The Luminous Flux from a monochromatic source producing light at a single wavelength is easiest to determine.

Φv = Φ *Vλ * (683 lm/W)

For instance, a 5 mW laser pointer using at a wavelength of 680 nm produces

.005 W * .017 * 683 lm/W = .058 lm

While a 5 mW laser pointer at 630 nm produces

.005 W * .265 * 683 lm/W = .905 lm, a significantly greater luminous flux.

You would do the same to get the watts. then think about the spot size you have for application.

If you want to be more precise. You would measure the spectrum, then measure the intensity at a particular wavelength with a specific bandwidth with a calibrated photodetector. then you can work it all out.

The instruments mentioned by Wenqing above are intended to make that process easier.

To make very accurate photometric measurements can be challenging.