Surface energy of a drop of liquid is 16mJ. When 8 of such identical drops coalesce to form a single drop

${\mathrm{U}}_{\mathrm{i}}=\mathrm{Initial}\mathrm{energy}=8\times 4{\mathrm{\pi r}}^2.\mathrm{T}=8\times 16\mathrm{mJ}$

If R be the radius of the bigger drop, Then

$\begin{array}{l}\frac{4}{3}{\mathrm{\pi R}}^3=8\times \frac{4}{3}{\mathrm{\pi r}}^3\Rightarrow \mathrm{R}=2\mathrm{r}\\ \therefore {\mathrm{U}}_{\mathrm{f}}=\mathrm{Final}\mathrm{surface}\mathrm{energy}=4{\mathrm{\pi R}}^2\mathrm{T}=4\times 4{\mathrm{\pi r}}^2\mathrm{T}=4\times 16\mathrm{mJ}\\ \therefore \mathrm{energy}\mathrm{released}={\mathrm{U}}_{\mathrm{i}}-{\mathrm{U}}_{\mathrm{f}}=(8\times 16-4\times 16)\mathrm{mJ}=64\mathrm{mJ}\end{array}$