1. Potential of an isolated conducting spherical drop

For a spherical conductor of radius $r$ carrying charge $q$, the potential is

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

So,

$q = \frac{Vr}{k}.$

2. Initial situation: 512 identical drops

Each small drop has potential $2\,\text{V}$. Let the radius of each small drop be $r$ and charge on each be $q$.

Thus,

$2 = k\frac{q}{r}.$

There are $512$ such drops.

3. When drops combine

If $n=512$ identical drops coalesce into one big drop:

• Total charge becomes $Q = 512q.$

• Volume is conserved: $\frac{4}{3}\pi R^3 = 512\left(\frac{4}{3}\pi r^3\right)$ $R^3 = 512r^3$ $R = 8r$ since $512 = 8^3$.

4. Potential of the big drop

The potential of the new drop is

$V' = k\frac{Q}{R} = k\frac{512q}{8r} = 64\,k\frac{q}{r}.$

But for each small drop,

$k\frac{q}{r} = 2.$

Therefore,

$V' = 64 \times 2 = 128\,\text{V}.$

5. Final answer

$\boxed{128}$

This also follows from the standard result:

$V' = n^{2/3}V$

for $n$ identical drops combining. Here,

$V' = 512^{2/3}\times 2 = 64\times 2 = 128\,\text{V}.$