If $n$ number of identical droplets of water, each of radius $r$, coalesce to form a single drop of radius $R$, the resulting rise in the temperature of water is given by (here $\rho$ is the density of water, s its specific heat and $\sigma$ its surface tension) 

Final energy = 4$\mathrm{\pi }$R² $\sigma$$={\mathrm{\pi D}}^2\mathrm{\sigma }$  where R is the radius of the big drop.

Initial energy = $\left(4{\mathrm{\pi r}}^2\right)n\sigma =\pi d^{2}n\sigma$, where d = 2r, is the radius of each tiny drop

Now ${\mathrm{R}}^3={\mathrm{nr}}^3$ or D³ = nd³ or $D=n^{1/3}d or d=D/n^{1/3}$ . 

Therefore, Initial energy= $\frac{{\mathrm{\pi D}}^2\mathrm{n\sigma }}{n^{2/3}}={\mathrm{\pi D}}^2{\mathrm{\sigma n}}^{1/3}$

 Hence, change in energy = ${\mathrm{\pi D}}^2{\mathrm{\sigma n}}^{1/3}-{\mathrm{\pi D}}^2\sigma ={\mathrm{\pi D}}^2\sigma \left(n^{1/3}-1\right)=4{\mathrm{\pi R}}^2({\mathrm{n}}^{1/3}-1)\mathrm{\sigma }$

Work done is

          $W=4{\mathrm{\pi R}}^2\left({\mathrm{n}}^{1/3}-1\right)\mathrm{\sigma }=4{\mathrm{\pi R}}^2\left(\frac{\mathrm{R}}{\mathrm{r}}-1\right)\sigma ----- \left(1\right) \left(\because \mathrm{R}={\mathrm{n}}^{1/3}\mathrm{r}\right)$

Heat produced  $Q=ms∆T=\frac{4}{3}{\mathrm{\pi R}}^3\mathrm{\rho s}∆\mathrm{T} \left(2\right)$

Equating (1) and (2), we find that choice (b) gives the correct expression for $∆$T.