Let radius of the small drop be $r$ and the radius of the big drop be $R$. Then, total volume will remain constant, therefore

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

$\Rightarrow R=10r$

Initial total surface energy, $E_1=1000\times 4\pi r^2\times S$ and 

$E_2=4\pi (10r{)}^2 S$

$$\begin{gathered} \Rightarrow \frac{E_1}{E_2}=\frac{10}{1} \\ \text{Hence}, x=10. \end{gathered}$$