(b) Given : $u = S \times 4\ p\ R^{2}$; when droplet is splitted into 1000 droplets each of radius $r$, then
$\frac{4}{3}\pi R^{3} = 1000 \times \frac{4}{3}\pi r^{3}$ or $r = R/10$
$\therefore$ Surface energy of all droplets
$= S \times 1000 \times 4\pi r^{2} = S \times 1000 \times 4\pi(R/10)^{2}$
$= 10\ (S\ 4\pi/R^{2}) = 10u$