1. Surface energy formula

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

For a spherical drop of radius $r$, $A = 4\pi r^2$ So, $E = T(4\pi r^2)$

2. Let radius of each small droplet be $r$

Then surface energy of one small droplet is $E_1 = 4\pi r^2 T$

For $1000$ small droplets, total surface energy is $E_s = 1000 \cdot 4\pi r^2 T$

3. Find radius of big drop

When $1000$ droplets coalesce, volume is conserved.

Volume of one small droplet: $V_1 = \frac{4}{3}\pi r^3$

Volume of $1000$ droplets: $1000 \cdot \frac{4}{3}\pi r^3$

Let radius of big drop be $R$. Then $\frac{4}{3}\pi R^3 = 1000 \cdot \frac{4}{3}\pi r^3$

So, $R^3 = 1000r^3$ $R = 10r$

4. Surface energy of big drop

$E_b = 4\pi R^2 T = 4\pi (10r)^2 T = 400\pi r^2 T$

5. Take the ratio

$\frac{\text{surface energy of 1000 droplets}}{\text{surface energy of big drop}} = \frac{1000\cdot 4\pi r^2 T}{400\pi r^2 T}$

$= \frac{4000}{400} = 10$

Given that this ratio is $\frac{10}{x}$ So, $\frac{10}{x} = 10$

Hence, $x = 1$

6. Final answer

$\boxed{1}$