What heat energy is required to completely boil 200 ml of pure water at 1 atmosphere of pressure

density of water at 20 °C = 1.00 g / ml
Cp (water) = 75.0 J K-1 mol-1
ΔHm(vap) = 40.7 kJ mol-1
Mm (H2O) = 18 g

It depends on what the initial temperature of the water is. I am assuming that it is 20°C since the density of water is given at this temperature (if not, just replace 20°C with the initial temperature).

Since 200 ml of water weighs (200)(1.00) = 200 g and 200g of water is 200/18 = 11.1 mol, we see that the energy required to warm the water up from 20°C to 100°C, the boiling point of water, is:
Q = mCΔT
= (11.1 mol)[75.0 J/(°C*mol)](100°C - 20°C)
= 6.66 x 10^5 J.

At this point, the energy required to boil the water is:
Q = m*H(v)
= (11.1 mol)(40700 J/mol)
= 4.518 x 10^5 J.

Thus, the total energy required is:
6.66 x 10^5 J + 4.518 x 10^5
= 1.12 x 10^6 J or 1120 kJ.

I hope this helps!