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

One drop of mercury has a radius of R. 

The volume of one drop is = $\frac{4}{3}{\mathrm{\pi R}}^3$

Total volume,

$V=2\times \frac{4}{3}{\mathrm{\pi R}}^3=\frac{8}{3}{\mathrm{\pi R}}^3$

Let R′ represent the large drop's radius.
The large drop's volume is also V.

$\frac{4}{3}\mathrm{\pi R}{\text{'}}^3=\frac{8}{3}{\mathrm{\pi R}}^3$

$\mathrm{R}{\text{'}}^3=2{\mathrm{R}}^3$

$\mathrm{R}\text{'}=2^{\frac{1}{3}}\mathrm{R}$

The surface area of the two drops,

${\mathrm{S}}_1=2\times 4{\mathrm{\pi R}}^2=8{\mathrm{\pi R}}^2$

The surface area of the resultant drop is,

${\mathrm{S}}_1=2\times 4\mathrm{\pi R}{\text{'}}^2=4\mathrm{\pi }{2}^{\frac{2}{3}}{\mathrm{R}}^2$

Let T represent the mercury's surface tension. Therefore, before coalescing, the surface energy of the two droplets is

${\mathrm{U}}_1={\mathrm{S}}_1\mathrm{T}=8{\mathrm{\pi R}}^2\mathrm{T}$

and surface energy after coalescence,

${\mathrm{U}}_2={\mathrm{S}}_2\mathrm{T}=2^{\frac{2}{3}}\times 4{\mathrm{\pi R}}^2\mathrm{T}$

$\frac{{\mathrm{U}}_1}{{\mathrm{U}}_2}=\frac{8{\mathrm{\pi R}}^2\mathrm{T}}{2^{\frac{2}{3}}\times 4{\mathrm{\pi R}}^2\mathrm{T}}=\frac{2}{2^{{\displaystyle \frac{2}{3}}}}=2^{\frac{1}{3}}$

Hence the correct answer is $2^{\frac{1}{3}}:1.$