a
Initial surface energy \(=0.45 \times 4 \pi\left(10^{-3}\right)^2\)

\(\frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text {new }}^3\)

\(10^{-3}=5 R_{\text {new }}\)

\(R_{\text {new }}=\frac{10^{-3}}{5}\,m\)

So, final surface energy \(=0.45 \times 125 \times 4 \pi\left(\frac{10^{-3}}{5}\right)^2\)

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

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

\(=2.26 \times 10^{-5}\,J\)