Volume of 8 small drops = Volume of big drop ∴(34πr3)×8=34πR3 ⇒2r=R ....(i)
According to charge conservation 8q=Q ....(ii)
Potential of one small drop (V′)=4πε0rq
Similarly, potential of big drop (V)=4πε0RQ
Now, VV′=Qq×rR ⇒20V′=8q9×r2r
[from Eqs. (i) and (ii)] ∴V′=5V