1. Surface energy of a liquid drop

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

For a spherical drop of radius $r$: $A = 4\pi r^2$ So, $U = 4\pi r^2 S$

2. Initial surface energy

The original drop has radius $R = 10^{-2}\,\text{m}$ Hence initial surface area is $A_i = 4\pi R^2$

So initial surface energy: $U_i = 4\pi R^2 S$

Given $S = \frac{0.1}{4\pi}\,\text{N m}^{-1}$ Thus $U_i = 4\pi R^2 \cdot \frac{0.1}{4\pi} = 0.1R^2$

Now substitute $R=10^{-2}$: $U_i = 0.1 \times (10^{-2})^2 = 0.1\times 10^{-4} = 10^{-5}\,\text{J}$

3. Relation between radii after division

Suppose the drop divides into $K$ identical drops each of radius $r$.

By conservation of volume: $\frac{4}{3}\pi R^3 = K\cdot \frac{4}{3}\pi r^3$ $R^3 = Kr^3$ $r = R K^{-1/3}$

4. Final surface energy

Total final surface area: $A_f = K\cdot 4\pi r^2 = K\cdot 4\pi (R^2K^{-2/3})$ $A_f = 4\pi R^2 K^{1/3}$

Hence final surface energy: $U_f = S A_f = 4\pi R^2 S\, K^{1/3}$

But $4\pi R^2S = U_i = 10^{-5}\,\text{J}$, so $U_f = 10^{-5} K^{1/3}$

5. Change in surface energy

Given increase in surface energy: $\Delta U = U_f - U_i = 10^{-3}\,\text{J}$

So, $10^{-5}(K^{1/3}-1)=10^{-3}$

Divide by $10^{-5}$: $K^{1/3}-1 = 100$ $K^{1/3} = 101$

Thus, $K = 101^3$

Now, $101^3 = 1030301 \approx 10^6$

Since $K = 10^\alpha$, $\alpha \approx 6$

6. Final answer

$\boxed{\alpha = 6}$

This matches the stored correct answer.