1. For an isolated spherical drop, electric potential is

$V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} = k\frac{q}{r}$

where $q$ is the charge and $r$ is the radius.

2. Let each small drop have:

• charge $q$
• radius $r$
• potential $V_1 = 22\,\text{V}$

So,

$V_1 = k\frac{q}{r} = 22$

3. When $27$ identical drops combine:

• total charge becomes

$Q = 27q$

• total volume becomes $27$ times, so radius becomes

$R = 27^{1/3}r = 3r$

4. Potential of the bigger drop:

$V_2 = k\frac{Q}{R} = k\frac{27q}{3r} = 9k\frac{q}{r}$

Using $k\frac{q}{r} = 22$,

$V_2 = 9 \times 22 = 198\,\text{V}$

5. Therefore, the potential of the bigger drop is

$\boxed{198}$