Answer
5600 kg of natural water per hour.
Work Step by Step
We are told to assume an efficiency of $33\%$. This means that the reactions actually generate a power of 1150MW/0.33.
Now that we know the average power for an hour, find the amount of energy.
$$\frac{(1150\times10^6 J)/0.33}{s}\frac{3600s}{hour}\frac{MeV}{1.60\times10^{-13}J}=7.841\times10^{25}MeV/h$$
Change this to an amount of deuterium. If we assume that the two d-d reactions take place at equal rates, then 4 deuterons react to release 4.03MeV+3.27 MeV= 7.30 MeV. This is an average of 1.825 MeV per deuteron.
$$(7.841\times10^{25}MeV/h)\frac{deuterium}{1.825MeV}=4.296\times10^{25} deuterons/h$$
Convert that figure to an amount of water. We know the fraction of deuterium vs regular H in natural water is $0.0115\%$.
$$(4.296\times10^{25} deuterons/h)\frac{1H}{0.000115d}\frac{H_2O}{2H}\frac{0.018kg}{6.02\times10^{23}molecules}$$
$$\approx5600\;kg/h$$
