The problem involves 64 identical drops, each charged up to a potential of $$10 \mathrm{~mV}$$, that are combined to form a bigger drop. We are asked to find the potential of the bigger drop.

The potential of each drop is given by:

$$V = \frac{Kq}{r}$$

where $$K$$ is the Coulomb constant, $$q$$ is the charge on each drop, and $$r$$ is the radius of each drop.

The radius of the bigger drop is:

$$R = 4r$$

since the 64 identical drops combine to form a bigger drop.

The total charge on the 64 identical drops is:

$$Q = 64q$$

The potential of the bigger drop is:

$$V_{bigger} = \frac{KQ}{R} = \frac{K(64q)}{4r} = 16 \frac{Kq}{r} = 16V = 16 \times 10 \mathrm{~mV} = 160 \mathrm{~mV}$$

Therefore, the potential of the bigger drop is $$160 \mathrm{~mV}$$.