The surface area of the liquid drop is $\mathit{A}=4{\mathrm{\pi R}}^2$

Its surface energy is $\mathit{E}$

When the drop splits in $512$ droplets, the surface area of each droplet is $4{\mathrm{\pi r}}^2$

$\therefore$ Total surface area ${\mathit{\text{A}}}_2$ $=512\times 4{\mathrm{\pi r}}^2$

The volume of bigger drop is $\frac{4}{3}{\mathrm{\pi R}}^3$ and volume of small droplets is $512\times \frac{4}{3}{\mathrm{\pi r}}^3$

$\therefore \frac{4}{3}{\mathrm{\pi R}}^3=512\times \frac{4}{3}{\mathrm{\pi r}}^3\Rightarrow \mathit{r}=\frac{\mathit{R}}{8}$

$\therefore {\mathit{A}}_2=512\times 4{\mathrm{\pi r}}^2=512\times 4\mathit{\pi }{\left(\frac{\mathit{R}}{8}\right)}^2=8{\mathit{A}}_1$

Surface energy $\mathit{E}=\mathit{A}.\mathit{T}$ ($\mathit{T}$ is surface tension and $\mathit{A}$ is area)

$\therefore \frac{{\mathit{E}}_{\mathit{n}}}{\mathit{E}}=\frac{{\mathit{A}}_2\cdot \mathit{T}}{{\mathit{A}}_1\cdot \mathit{T}}=\frac{8\cdot {\mathit{A}}_1}{{\mathit{A}}_1}=8$

$\therefore \mathit{ }{\mathit{E}}_{\mathit{n}}=8\mathit{E}$