Explanation

(c) Let r be the radius of one droplet .
$Now,\ \ \ \ \frac{4}{3}\pi R^{3} = 10^{6} \times \frac{4}{3}\pi r^{3}$
$\ \ \ \ \ r = \frac{R}{100} = \frac{1}{100}\ cm = 10^{- 4}\ m$
$\ \ \ A_{i} = 4\pi R^{2}$
$\ \ \ A_{f} = 10^{6} \times 4\pi r^{2}$
Change in area,
$\ \ \mathrm{\Delta}A = A_{f} - A_{i} = 4\pi \times 99 \times 10^{- 4}\ m^{2}$
Increase in surface energy
$\ \ = S\ \mathrm{\Delta}A = 32 \times 10^{- 2} \times 4\pi \times 99 \times 10^{- 4}\ J$
$\ = 3.98 \times 10^{- 2}J\ $
The increase in surface energy is on the expense of internal energy, so energy expended $= 3.98 \times 10^{- 2\ }J\ $