As given that,
Area of square sheet
$$
(A)=\left(10^{-2}\right) \mathrm{m}^2=10^{-2} \times 10^{-2} \mathrm{~m}^2=10^{-4} \mathrm{~m}^2
$$
Thickness $(\mathrm{d})=10^{-3} \mathrm{~m}$
Current (i) $=100 \times 10^{-6} \mathrm{~A}=10^{-4} \mathrm{~A}$
Intensity $(\mathrm{I})=100 \mathrm{~W} / \mathrm{m}^2$
$\Rightarrow \lambda=660 \mathrm{~nm}=660 \times 10^{-9} \mathrm{~m}$
$\rho_{\mathrm{Na}}=0.97 \mathrm{~kg} / \mathrm{m}^3$
Avogadro's number $=6 \times 10^{26} \mathrm{~kg}$ atom
Volume of sodium target $=\mathrm{A} \times \mathrm{d}$
$$
=10^{-4} \times 10^{-3}=10^{-7} \mathrm{~m}^3
$$
We know that, $6 \times 10^{26}$ atoms of $\mathrm{Na}$ weights $=23 \mathrm{~kg}$
So, volume of $6 \times 10^6$ Na atoms $=\frac{23}{0.97} \mathrm{~m}^3$
volume occupied by one Na atom
$$
=\frac{23}{0.97 \times\left(6 \times 10^{26}\right)}=3.95 \times 10^{-26} \mathrm{~m}^3
$$
Number of atoms in (Na) target
$$
=\frac{10^{-7}}{3.95 \times 10^{-26}}=2.53 \times 10^{18} \text { Atoms, }
$$
Let $\mathrm{n}$ be the number of photons falling per second on the target,
Energy of each photon $=\operatorname{nh} v \quad\left(\because \mathrm{v}=\frac{\mathrm{c}}{\lambda}\right)$
Total energy falling per second on target $=\frac{n h c}{\lambda}=$ IA $($ Intensity $\times$ Area)
$$
\begin{aligned}
&\therefore \quad \mathrm{n}=\frac{\mathrm{IA} \lambda}{\mathrm{hc}} \\
&\Rightarrow \quad=\frac{100 \times 10^{-4} \times\left(660 \times 10^{-9}\right)}{\left(6.62 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}=3.3 \times 10^{16}
\end{aligned}
$$
Let $\mathrm{P}$ be the probability of emission of photo electrons per atom per photon.
The number of photoelectrons emitted per second
$$
\begin{aligned}
N &=P . n(\text { No. of sodiuma atom }) \\
N &=P \times n \times(N a) \\
&=P \times\left(3.3 \times 10^{16}\right) \times\left(2.53 \times 10^{18}\right)
\end{aligned}
$$
Now, according to question,
$$
\mathrm{i}=100 \mu \mathrm{A}=100 \times 10^{-6}=10^{-4} \mathrm{~A} \text {. }
$$
Current, $\mathrm{i}=\mathrm{Ne}$
Put $\mathrm{i}$ and $\mathrm{N}$ value in the equation
$$
\begin{aligned}
&\therefore 10^{-4}=\mathrm{P} \times\left(3.3 \times 10^{16}\right) \times\left(2.53 \times 10^{18}\right) \\
&\times\left(1.6 \times 10^{-19}\right) \\
&\Rightarrow \mathrm{P}=\frac{10^{-4}}{\left(3.3 \times 10^{16}\right) \times\left(2.53 \times 100^{18}\right)_{\times}\left(1.6 \times 10^{-19}\right)} \\
&=7.48 \times 10^{-21} \simeq 7.5 \times 10^{-21} \\
&
\end{aligned}
$$
So that, 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.