Hint: The no. of photons emitting from a surface depends on the intensity of radiation.

Step 1: Find the no. of photons emitted by the bulb per second.

Given,

$\mathrm{P}=20\mathrm{W}, \mathrm{\lambda }=500\stackrel{°}{\mathrm{A} }=5000 \times {10}^{-10} \mathrm{m}$
$\mathrm{d}=2 \mathrm{m}, {\mathrm{\varphi }}_0=2\mathrm{eV}, \mathrm{r} =1.5 \stackrel{\mathrm{o}}{\mathrm{A}}=1.5\times {10}^{-10 }\mathrm{m}$

(i) Number of photons emitted by bulb per second is;

$\mathrm{n}=\frac{\mathrm{p}}{\mathrm{hc}/\mathrm{\lambda }}=\frac{\mathrm{p\lambda }}{\mathrm{hc}}$
$=\frac{20\times (5000\times {10}^{-10})}{(6.62\times {10}^{-34})\times (3\times {10}^8)}$
$=5\times {10}^{18}$

Step 2: Find the energy of incident photons.

(ii) Energy of the incident photon,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}=\frac{(662\times {10}^{-34})(3\times {10}^8)}{5000\times {10}^{-10}\times 1.6\times {10}^{-19}}$
$=2.48 \mathrm{eV}$

As this energy is greater than 2 eV (i.e., work of metal surface), hence photoelectric emission takes place. 

Step 3: Find the time of getting energy.

(iii) Let $∆$t be the time spent in getting the energy, $\varphi$=work function of the metal.

Consider the figure, 

  $\begin{array}{rl}\frac{\mathrm{P}}{4{\mathrm{\pi d}}^2}\times {\mathrm{\pi r}}^2\mathrm{\Delta t}& ={\mathrm{\varphi }}_0\end{array}$
$\begin{array}{rl}\Rightarrow \mathrm{\Delta t}& =\frac{4{\mathrm{\varphi }}_0{\mathrm{d}}^2}{{\mathrm{Pr}}^2}=\frac{4\times (2\times 1.6\times {10}^{-19})\times 2^2}{20\times {(1.5\times {10}^{-10})}^2}\approx 28.4 \sec \end{array}$

Step 4: Find the no. of photon received in $∆$t.

(iv) Number of photons received by the atomic disc in time $∆$t is:

$\begin{array}{rl}\mathrm{N}& =\frac{{\mathrm{n}}^{\mathrm{\prime }}\times {\mathrm{\pi r}}^2}{4{\mathrm{\pi d}}^2}\times \mathrm{\Delta t}\end{array}$
$\Rightarrow \mathrm{N} =\frac{{\mathrm{n}}^{\mathrm{\prime }}{\mathrm{r}}^2\mathrm{\Delta t}}{4{\mathrm{d}}^2}$
$=\frac{(5\times {10}^{19})\times {(1.5\times {10}^{-10})}^2\times 28.4}{4\times (2{)}^2}=2$

(v) As the time of emission of electrons is 11.04 is s. 

Hence, the photoelectric emission is not instantaneous in this problem. 

ln photoelectric emission, there is a collision between the incident photon and free electron of the metal surface, which lasts for a very-very short interval of time $(={10}^{-9}s)$, hence we say photoelectric emission is instantaneous.