1. Surface energy of a liquid drop

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

For a spherical drop of radius $r$, $A = 4\pi r^2$ So surface energy of one drop is $E_1 = 4\pi r^2 T$

Since there are two identical drops, initial surface energy is $E_i = 2 \times 4\pi r^2 T = 8\pi r^2 T$

2. Find radius of bigger drop after coalescence

When two drops coalesce, volume is conserved.

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

Volume of two drops: $2\cdot \frac{4}{3}\pi r^3$

Let radius of bigger drop be $R$. Then $\frac{4}{3}\pi R^3 = 2\cdot \frac{4}{3}\pi r^3$

Cancelling common factors, $R^3 = 2r^3$ $R = 2^{1/3}r$

3. Final surface energy

Surface area of bigger drop: $A_f = 4\pi R^2 = 4\pi (2^{1/3}r)^2 = 4\pi 2^{2/3} r^2$

Hence final surface energy is $E_f = T \cdot A_f = 4\pi r^2 T\,2^{2/3}$

4. Surface energy released

Released energy = decrease in surface energy: $\Delta E = E_i - E_f$

So, $\Delta E = 8\pi r^2 T - 4\pi r^2 T\,2^{2/3}$

Factor out $4\pi r^2 T$: $\Delta E = 4\pi r^2 T\left(2 - 2^{2/3}\right)$

5. Match with options

This matches: $\boxed{4\pi r^2 T\left(2-2^{2/3}\right)}$ So the correct option is C.