Sixty four conducting drops each of radius 0 . 02m and each...

Total volume will be constant. Therefore, $n\left(\dfrac{4\pi}{3}{r}^{3}\right) = \dfrac{4\pi}{3}{R}^{3} ⇒ 64\dfrac{1}{3}r = R ⇒ R = 4r$

Final surface charge density ${\sigma}^{'} = \dfrac{n{\sigma}_{0}4\pi{r}^{2}}{4\pi{R}^{2}} = \dfrac{64 \times {\sigma}_{0}{r}^{2}}{16{r}^{2}} ⇒ \dfrac{{\sigma}^{'}}{{\sigma}_{0}} = \dfrac{4}{1}$