I think this question is quite tricky. It's hard to say it's gravitational potential or electric potential. We need a bit of analysis:
Rain is usually not pure water, i.e. it's electrically conducting. From the time the rain drop is formed to the time it falls through a long distance, it accumulates an amount of charges due to "rubbing" with the surrounding. So, we can view the rain drop as a conducting charged sphere. Assume that V is electric potential and the rain drop can be treated as isolated (so that V is due to ONLY the rain drop itself, no external effect by other rain drops). Let Q and R denote the charge and the radius of the rain drop. You can calculate V in term of Q and R, can't you?
Then when 2 identical rain drops coalesce, it forms a bigger rain drop of charge 2Q. The radius of the new rain drop can be calculated (the total volume is conserved). Again, the electric potential of this new rain drop V' can be calculated in term of Q and R, as it's an isolated conducting spherical rain drop. Finally find the ratio of V' and V, and thus, express V' in term of V.
So what if V is gravitational potential? My best guess is, it's quite pointless or way too easy if so. For 2 identical rain drops to coalesce, they must be at the same height. The newly formed rain drop should be at that height as well, and thus, its potential is 2V (as mass is doubled). Too obvious! I have just only considered the gravitational potential due to the gravitational field of the Earth. But taking into account the gravitational potential due to gravitational field of the rain drop itself is again quite pointless, and more importantly, impossible to lead to a uniform potential at every point in the rain drop.
