$r=$ radius of smaller drop

Volume will remain same

$\frac{4}{3} \pi R ^{3}=64 \times \frac{4}{3} \pi r ^{3}$

$R =4 r$

$Q =64 q ;$

$q$ : charge of smaller drop

$Q$ : Charge of combined drop

$\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{\frac{ Q }{4 \pi R ^{2}}}{\frac{ q }{4 \pi r ^{2}}}=\frac{ Q }{ q } \cdot \frac{ r ^{2}}{ R ^{2}}$

$=64 \frac{ r ^{2}}{16 r ^{2}}=4$

$\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{4}{1}$