Let $R$ be the radius of bigger drop and $r$ be the radius of smaller drop.

Here, volume will remain conserved, so $V_R=V_r$.

$\Rightarrow \frac{4}{3}\pi R^3=1000\times \frac{4}{3}\pi r^3$

$\Rightarrow R=10r$

As surface energy is stated as $u=SA$, here, $S$ is surface tension and $A$ is surface area.

Thus, $u_i=S\left(4\pi R^2\right)$ and $u_f=S\left(4\pi r^2\right)\times 1000$

The ratio is $\frac{u_f}{u_i}=\frac{1000r^2}{R^2}$

$\Rightarrow \frac{u_f}{u_i}=\frac{10}{1}$

Hence, the value of $x=1$.