Voltage of each mercury drop \\\\(= 10 V\\\\)\\\\(\\Rightarrow \\dfrac {kq}{r}= 10 V\\\\)\\\\(\\Rightarrow q= \\dfrac {10r}{k} \\\\) \\\\(total \\ charge =\\dfrac {27\ imes 10r}{k}\\\\)Equating volumes to find radius of bigger charge, \\\\(\\dfrac {4}{3}\\pi (R^3)=27\\cdot \\dfrac {4}{3}\\pi r^3\\\\)\\\\(R=3r\\\\)final voltage\\\\( = \\dfrac {kq}{R}=\\dfrac {k \ imes 27\ imes 10r}{k\ imes 3r}\\\\)\\\\(=90 V\\\\)