For the following, let us assume you:

1) have your full beam going through the sample (fulfilled for standard cuvette setups)

2) have a reliable absolute power measurement, not just a relative one (not the case for all instruments, check detector specifications)

In that case you can just divide your integrated absorbed overall energy by h*nu and you have your photon number.

The only part that's really crucial here is fulfilling the second condition, the rest is pretty standard.

-'the rest is pretty standard' - as Jürgen stated above.

I have got the impression from your question that in the course of time of your basic studies you did not really have got in touch with the Beer-Lambert (often also called Lambert-Beer) law of attenuation.

It is a relation which connects the amount of primary radiation I0 to the transmitted radiation I ( i. e. the amount of radiation which passes and leaves the sample) via an exponential expression such as:

I = I0*exp(-µ*d); with µ as the linear attenuation coefficient and d the net path length in the sample;

or more often used in the optical regime:

I= Io*10-A; with A named as the absorbance.

The absorbance A is related to the the path length d, the concentration c and the molar absorptivity (sometimes also named as the molar absorption coefficient) epsilon via:

A = epsilon*c*d.

So we have : I= Io*10_A = I0*10-(epsilon*c*d)

You asked for the amount of light Ia which is absorbed in the sample:

Ia = I0 -I = I0 - I0*10-A = I0*(1-10-A) =I0*[1-10-(epsilon*c*d)]

For examples dealing with the Beer-Lambert law please have a look for example at:

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Electronic_Spectroscopy_Basics/The_Beer-Lambert_Law

Thanks to all. I got the answer