The surface area of the liquid drop is $A=4\pi R^2$
Its surface energy is $E$
When the drop splits in $512$ droplets, the surface area of each droplet is $4\pi r^2$
$\therefore$ Total surface area ${\mathit{\text{A}}}_2$ $=512\times 4\pi r^2$
The volume of the bigger drop is $\frac{4}{3}\pi R^3$ and the volume of small droplets is $512\times \frac{4}{3}\pi r^3$
$\therefore \frac{4}{3}\pi R^3=512\times \frac{4}{3}\pi r^3\Rightarrow r=\frac{R}{8}$
$\therefore A_2=512\times 4\pi r^2=512\times 4\mathrm{\pi }{\left(\frac{R}{8}\right)}^2=8A_1$
Surface energy $E=A.T$ ($T$ is surface tension and $A$ is area)
$\therefore \frac{E_{\mathrm{n}}}{E_0}=\frac{A_2\cdot \mathrm{T}}{A_1\cdot \mathrm{T}}=\frac{8\cdot A_1}{A_1}=8$
$\therefore \mathrm{ }{E}_{\mathrm{n}}=8E$