Answer
See answers.
Work Step by Step
In the d-d reaction, 2 deuterons react to release 4.03MeV. We know the fraction of deuterium vs regular H in natural water is $0.0115\%$.
Find the energy in a kilogram of water.
$$(1.00\;kg\;H_2O)\frac{4.03MeV}{2\;deuterium}\frac{0.000115d}{1H}\frac{2H}{H_2O}\frac{6.02\times10^{23}molecules}{0.018kg}$$
$$=1.550\times10^{22}MeV$$
This equals $2.48\times10^9J$.
Compared to the energy released by burning 1.0 kg of gasoline, this is greater by a factor of 50.
$$\frac{2.48\times10^9J}{5\times10^7J}\approx 50$$
