Two small drops of mercury each of radius R coalesce to form a...

$\dfrac{4}{3}\pi{R}^{3} + \dfrac{4}{3}\pi{R}^{3} = \dfrac{4}{3}\pi{R}^{' 3}$

${R}^{'} = 2\dfrac{1}{3}R \ldots \left(i\right)$

${A}_{i} = 2\left[4\pi{R}^{2}\right]$

${A}_{f} = 4\pi{R}^{' 2}$

$\dfrac{{U}_{i}}{{U}_{f}} = \dfrac{{A}_{i}}{{A}_{f}} = \dfrac{2{R}^{2}}{{2}^{\dfrac{2}{3}}{R}^{2}} = 2\dfrac{1}{3}$