Lets say radius of small droplets is $r$ and that of big drop is $R$. Total volume of the drops will remains unchanged. Therefore, 

$\frac{4}{3}\pi R^3=1000\left(\frac{4}{3}\pi r^3\right)$

$\Rightarrow R=10r$

Now, initial surface energy is, $U_i=1000\left(4\pi r^2 S\right)$ and the final surface energy is, 

$U_f=4\pi R^2 S$

$=100\left(4\pi r^{2}S\right)$

Clearly, $U_f=\frac{1}{10}{U}_i$.