1. Surface energy is proportional to surface area

For a liquid drop, $u = T \cdot A$ where $T$ is surface tension and $A$ is surface area.

So, $\frac{u_f}{u_i} = \frac{A_f}{A_i}$

since the liquid is the same, $T$ remains same.

2. Let the radius of the original drop be $R$

Then its volume is $V_i = \frac{4}{3}\pi R^3$

It splits into $1000$ identical drops, each of radius $r$.

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

Cancelling common factors, $1000r^3 = R^3$

$r = \frac{R}{10}$

because $1000 = 10^3$.

3. Initial surface area

$A_i = 4\pi R^2$

4. Final total surface area

Surface area of one small drop: $4\pi r^2$

For $1000$ drops, $A_f = 1000 \cdot 4\pi r^2$

Substitute $r = R/10$: $A_f = 1000 \cdot 4\pi \left(\frac{R}{10}\right)^2$

$A_f = 1000 \cdot 4\pi \frac{R^2}{100}$

$A_f = 10 \cdot 4\pi R^2$

So, $A_f = 10A_i$

Hence, $\frac{u_f}{u_i} = \frac{A_f}{A_i} = 10$

5. Compare with given form

Given, $\frac{u_f}{u_i} = \frac{10}{x}$

So, $10 = \frac{10}{x}$

Therefore, $x=1$

Final Answer: $\boxed{1}$