Two small drops of mercury each of radius r form a single large drop. The ratio of surface energy before and after this change is

Suppose R be the radius of bigger drop. Then, by equating volumes, we get

$2\left(4/3\pi r^3\right)=(4/3)\pi R^3\text{ or }R=(2{)}^{1/3}r$

Now, surface energy ex surface area

$\therefore \frac{U_1}{U_2}=\frac{A_1}{A_2}=\frac{2\cdot r^2}{R^2}=\frac{2}{2^{2/3}}$