Let a spherical drop of liquid of radius R divides itself into K spherical drops, each of radius r. The density of liquid does not change in this process. Hence, conservation of mass gives

$${4 \over 3}\pi {R^3} = {4 \over 3}\pi {r^3}K$$, i.e., $$r = R{K^{ - 1/3}}$$. ...... (1)

The surface energy of a drop of radius R is $${U_i} = 4\pi {R^2}S$$ and total surface energy of K drops of radius r is $${U_f} = 4\pi {r^2}KS$$. Note that the surface energy increases when a bigger drop is divided into multiple smaller drops. Total change in surface energy is

$$\Delta U = {U_f} - {U_i} = 4\pi S(K{r^2} - {R^2})$$ ...... (2)

Eliminate r from equations (1) and (2) to get

$$K = {\left( {{{\Delta U} \over {4\pi S{R^2}}} + 1} \right)^3}$$.

Substitute $$\Delta$$U = 10^$$-$$3 J, S = $${{0.1} \over {4\pi }}$$ N/m and R = 10^$$-$$2 m to get $$K = {(100 + 1)^3} \approx {10^6}$$.