A mercury drop of radius  ${10}^{-3}m$   is broken into 125 equal size droplets. Surface tension of mercury is $0.45N{m}^{-1}$. The gain in surface energy is 

Initial surface energy =  $0.45\times 4\pi {\left({10}^{-3}\right)}^2$
$\frac{4}{3}\pi {\left({10}^{-3}\right)}^2=125\times \frac{4\pi }{3}{R}_{new}^2 ={10}^{-3}=5R_{new}{R}_{new}=\frac{{10}^{-3}}{5}n$   
Final surface energy  = $0.45\times 125\times 4\pi {\left(\frac{{10}^{-3}}{5}\right)}^2$
Increase in energy =  $0.45\times 4\pi \times {\left({10}^{-3}\right)}^2[\frac{125}{25}-1]$  

$=4\times 0.45\times 4\pi \times {10}^{-6}=2.26\times {10}^{-5}J$