Eight mercury droplets having a radius of 1 mm and a charge of 0.066…

Since Volume remains constant so

Initial volume = Final Volume

$ 8\times \frac{4\pi}{3} r^3 = \frac{4\pi}{3}R^3$

$ \Rightarrow R = 2r = 2mm ( r = 1mm)$

Since charge also remain constant . it is same on both the drops.

$ \Rightarrow V = \frac{Q}{4\pi \epsilon_0 R} = \frac{9\times 10^9 \times 8\times 0.066\times 10^{-12}}{2\times 10^{-3}} = 2.4 V$