Explanation

(c) Volume remains constant after coalescing.
Thus,
$\frac{4}{3}\pi R^{3} = 2 \times \frac{4}{3}\pi r^{3}$
Where R is radius of bigger drop and r is radius of each smaller drop.
$\therefore\ R = 2^{\frac{1}{3}}r$
Now, surface energy per unit surface area is the surface tension.
So, surface energy, $W = T\mathrm{\Delta}A$
$or\ \ \ W = 4\pi r^{2}T$
Therefore, surface energy of bigger drop
$\ W_{1} = 4\pi(2^{\frac{1}{3}}r)^{2}T = \ (2^{\frac{2}{3}})4\pi r^{2}T$
Surface energy of smaller drop
$\ \ W_{2} = 4\pi r^{2}T$
Hence, required ratio
$\ \ \frac{W_{1}}{W_{2}} = 2^{2/3}:1$