As the volume of the mercury will remain same, therefore

$$\begin{gathered} \frac{4}{3}\pi R^3=125\left(\frac{4}{3}\pi r^3\right) \\ \Rightarrow R=5r \end{gathered}$$

Now, increase in surface energy

$∆U=125\times \left(S\times 4\pi r^2\right)-\left(S\times 4\pi R^2\right)$

 $=0.45\times 4\mathrm{\pi }\times {\left({10}^{-3}\right)}^2\left[\frac{125}{25}-1\right]$

$=4\times 0.45\times 4\pi \times {10}^{-6}$

$=2.26\times {10}^{-5}\mathrm{J}$