$2 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$

$R=(2)^{1 / 3} r$

We know that the terminal velocity of each individual drop is given as $v=\frac{\left(\frac{2}{9}\right) r^{2} g(\rho-\sigma)}{\eta}$

Thus the terminal velocity of the coalesced drop is given as $v^{\prime}=\frac{\left(\frac{2}{9}\right)\left(2^{1 / 3} r\right)^{2} g(\rho-\sigma)}{\eta}$

$v^{\prime}=2^{2 / 3} v=4^{1 / 3} v$