Hint: The intensity of the radiation gives the energy falling on the metal per second.

Step 1: Find the no. of atoms in the Na-target.

Given,

$\begin{array}{rl}\Rightarrow A& ={\left({10}^{-2}\mathrm{m}\right)}^2={10}^{-2}\times {10}^{-2}{\mathrm{m}}^2\\ & ={10}^{-4}{\mathrm{m}}^2\\ d& ={10}^{-3}\mathrm{m}\\ i& =100\times {10}^{-6}\mathrm{A}={10}^{-4}\mathrm{A}\\ Intensity , I& =100\mathrm{W}/{\mathrm{m}}^2\\ \Rightarrow \lambda & =660\mathrm{nm}=660\times {10}^{-9}\mathrm{m}\\ {\mathrm{\rho }}_{\mathrm{Na}}& =0.97 kg/m^3\end{array}$

Avogadro's number=$6\times {10}^{26} kg atom$

The volume of sodium target =A$\times d$

                   $={10}^{-4}\times {10}^{-3}$
$={10}^{-7} m^3$

We know that $6\times {10}^{26}$ atoms of Na weigh = 23 kg 

So, volume $6\times {10}^{26}$ Na atoms =$\frac{23}{0.97} m^3$

The volume occupied by one Na-atom$=\frac{23}{0.97\times (6\times {10}^{26})}=3.95\times {10}^{-26} m^3$

Number of Na-atoms in the target $(n_{Na})$$=\frac{{10}^{-7}}{3.95\times {10}^{-26}}=2.53\times {10}^{18}$

Step 2: Find the no. of photons falling on the target per sec.

Let n be the number of photons falling per second on the target. 

The energy of each photon=hc/$\lambda$

$Total energy falling per second on target =\frac{nhc}{\lambda }=lA$
$\Rightarrow n=\frac{IA\lambda }{hc}$
$\Rightarrow n=\frac{100\times {10}^{-4}\times (660\times {10}^{-9})}{(662\times {10}^{-34}\times (3\times {10}^8)}=3.3\times {10}^{16}$

Step 3: Find the probability of the emission of electrons.

Let P be the probability of emission per atom-photon. 

The number of photoelectrons emitted per second 

                $N=P\times n\times \left(n_{Na}\right)$
$=P\times (3.3\times {10}^{16})\times 2.53\times {10}^{18})$

Now, according to the question, 

             $i=100 \mu A =100\times {10}^{-6}={10}^{-4} A$

$\begin{array}{rl}Current, i& =N\mathrm{e}\\ {10}^{-4}& =P\times (3.3\times {10}^{16})\times (2.53\times {10}^{18})\times (1.6\times {10}^{-19})\\ \Rightarrow P& =\frac{{10}^{-4}}{(3.3\times {10}^{16})\times (2.53\times {10}^{18})\times (1.6\times {10}^{-19})}\\ & =7.48\times {10}^{-21}\end{array}$

Thus, the probability of emission by a single photon on a single atom is very much less than 1. It is due to this reason, the absorption of two photons by an atom is negligible.