Answer
$1.65\times10^{21}$ $photons/s.m^2$
Work Step by Step
If $P$ be the power of the laser, and $E$ be the energy of a single incident photon, the rate of incident photon can be expressed as:
$R_{in}=\frac{P}{E}$ ................$(1)$
For light of frequency $f$ and wavelength $\lambda$, the photon energy is
$E =hf$
or, $E=\frac{ch}{\lambda}$,
in which $c$ is the speed of light in vacuum, and $h$ is the Planck's constant.
In our case, the wavelength of the incident x-ray is $\lambda=633$ $nm$ $=633\times 10^{-9}$ $m$, $P=5.00$ $mW$ $=5.00\times 10^{-3}$ $W$
Therefore, the energy of a single the incident photon is
$E=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{633\times 10^{-9}}$ $J$
or, $E\approx3.142\times 10^{-19}$ $J$
$\therefore$ Using eq. $1$, the rate of incident photon is given by
$R_{in}=\frac{5.00\times 10^{-3}}{3.142\times 10^{-19}}$ $photons/s$
or, $R_{in}\approx 1.59\times10^{16}$ $photons/s$
The detector in the beam’s path totally absorbs the beam.
Therefore, the rate of photon absorbed by the detector is given by
$R_{abs}=R_{in}=1.59\times10^{16}$ $photons/s$
The diameter of the beam, $d=3.5$ $mm$ $=3.5\times 10^{-3}$ $m$. Therefore the an absorbing area of the director is
$A_{det}=\pi (\frac{3.5\times 10^{-3}}{2})^{2}$
$A_{det}=9.62 \times 10^{-6}$ $m^2$
$\therefore$ The rate of photon absorbed by per unit area of the detector is
$R=\frac{R_{abs}}{A_{det}}$
or, $R=\frac{1.59\times10^{16}}{9.62 \times 10^{-6}}$ $photons/s.m^2$
or, $R\approx 1.65\times10^{21}$ $photons/s.m^2$
