1. Potential energy of an isolated spherical conductor

For a spherical conductor of capacitance $C$, potential $V$, and charge $Q$:

$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$

For a sphere of radius $r$:

$C = 4\pi\varepsilon_0 r$

So for one small mercury drop at potential $V=10\,\text{V}$,

$U_s = \frac{1}{2} C_s V^2$

where

$C_s = 4\pi\varepsilon_0 r$

2. When 27 identical drops combine

If $27$ similar spherical drops combine, volume is conserved:

$\frac{4}{3}\pi R^3 = 27\left(\frac{4}{3}\pi r^3\right)$

Thus,

$R^3 = 27r^3 \implies R = 3r$

Hence capacitance of the big drop is

$C_b = 4\pi\varepsilon_0 R = 3C_s$

3. Charge on each small drop

Each small drop is at potential $10\,\text{V}$, so charge on each drop is

$q = C_s V$

Total charge on the combined drop:

$Q = 27q = 27C_sV$

4. Potential of the big drop

Potential of the big drop is

$V_b = \frac{Q}{C_b} = \frac{27C_sV}{3C_s} = 9V$

So,

$V_b = 9 \times 10 = 90\,\text{V}$

5. Potential energy of the big drop

$U_b = \frac{1}{2} C_b V_b^2$

Substitute $C_b = 3C_s$ and $V_b = 9V$:

$U_b = \frac{1}{2}(3C_s)(9V)^2$

$U_b = \frac{1}{2}(3C_s)(81V^2) = 243\left(\frac{1}{2}C_sV^2\right)$

But

$\frac{1}{2}C_sV^2 = U_s$

Therefore,

$U_b = 243U_s$

6. Required ratio

The potential energy of the bigger drop is

$\boxed{243}$

times that of a smaller drop.