Volume of 8 small drops = Volume of big drop
$\therefore \left(\frac{4}{3}\pi r^{3}\right) \times8 = \frac{4}{3}\pi R^{3}$
$ \Rightarrow 2r =R$ ....(i)
According to charge conservation
$8q = Q$ ....(ii)
Potential of one small drop $ \left(V'\right) = \frac{q}{4\pi\varepsilon_{0}r}$
Similarly, potential of big drop $ \left(V\right)= \frac{Q}{4\pi\varepsilon_{0}R}$
Now, $ \frac{V'}{V} = \frac{q}{Q}\times\frac{R}{r} $
$\Rightarrow \frac{V'}{20} = \frac{9}{8q}\times\frac{2r}{r}$
[from Eqs. (i) and (ii)]
$ \therefore V' = 5\,V $