Two small drops of mercury each of radius $$R$$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is:

Let each small drop have radius $$R$$. The surface energy of each drop is $$4\pi R^2 T$$, so the total surface energy before coalescence is $$E_i = 2 \times 4\pi R^2 T = 8\pi R^2 T$$.

After coalescence, volume is conserved: $$2 \times \frac{4}{3}\pi R^3 = \frac{4}{3}\pi r^3$$, giving $$r^3 = 2R^3$$, so $$r = 2^{1/3}R$$.

The surface energy of the large drop is $$E_f = 4\pi r^2 T = 4\pi (2^{1/3}R)^2 T = 4\pi \cdot 2^{2/3} R^2 T$$.

The ratio of initial to final surface energy is $$\frac{E_i}{E_f} = \frac{8\pi R^2 T}{4\pi \cdot 2^{2/3} R^2 T} = \frac{8}{4 \cdot 2^{2/3}} = \frac{2}{2^{2/3}} = 2^{1-2/3} = 2^{1/3}$$.

Therefore the ratio of total surface energy before to after is $$2^{1/3} : 1$$.