Let charge and radius of smaller drop is q and r respectively

For smaller drop, V = ${{kq} \over r} = 220\,V$

Let R be radius of bigger drop,

As volume remains the same

$\left( {{4 \over 3}\pi {r^3}} \right) \times 27 = {4 \over 3}\pi {R^3}$

$\Rightarrow$ R = 3r

Now, using charge conservation,

Q = 27q

V_{big drop} = ${{kQ} \over r} = {{k\left( {27q} \right)} \over r} = 9\left( {{{kq} \over r}} \right)$

= 9 $\times$ 220 = 1980 V