The volume of a sphere is given by $\frac{4}{3}\pi R^3$.

So the volume of the two small mercury drops each of radius $R$ is $2\times \frac{4}{3}\pi R^3$.

When they coalesce to form a larger drop, the volume is conserved. So, the volume of the larger drop is also $2\times \frac{4}{3}\pi R^3$.

Let's denote the radius of this large drop as $R'$.

Therefore, $2\times \frac{4}{3}\pi R^3 = \frac{4}{3}\pi {R'}^3$.

Solving for $R'$, we get $R' = 2^{1/3}R$.

Now, the surface energy of a sphere is proportional to its surface area, and the surface area of a sphere is given by $4\pi R^2$.

So, the ratio of total surface energy before and after the change is:

$$\frac{{2\times 4\pi R^2}}{{4\pi (2^{1/3}R)^2}} = \frac{{2}}{{2^{2/3}}} = {2^{1/3}}:1$$