0.287% of the water needs to evaporate.
The Enthalpy of vaporization for water is 40.65 kJ/mol, and the heat capacity of liquid water is 75.327 J/(mol*K). I could have used other values such as 4.1813 J/(g*K) for the specific heat of water, but then I'd need to either convert from moles to grams for the Enthalpy of vaporization, or convert from grams to moles for the specific heat of water. It's just easier if both constants use the same unit for the mass of material.
We really don't care how large the pool is, nor how hot the sun is, or even how warm the air is. So, I'm going to use a really small pool that only has 1 mole of water in it and I will assume that any temperature increase is from the sun. So how much energy is needed to raise 1 mole of water 1.55 degrees?
1 mol * 75.327 J/(mol*K) * 1.55 = 116.75685 J
So the small pool absorbed 116.75685 Joules of energy. Now to figure out how many moles of water needs to evaporate to carry off that energy.
116.75685 J * 0.001 kJ/J / 40.65 kJ/mol = 0.002872247 mol
So 0.002872247 mol of water must evaporate to carry off the heat from my tiny little pool, and conveniently since pool has only 1 mole of water to start with, that means that 0.287% of the water from the pool has to evaporate.
Note: I could use a larger pool, but all that will do is make the intermediate numbers larger, but won't affect the final result. And I'm lazy, so if you want to prove that to yourself, go ahead. But I'm done.
