Answer
0.39 grams of deuterium fuel.
Work Step by Step
If 960 watts is the average power for a year, find the amount of energy.
$$\frac{960J}{s}\frac{3.156\times10^7s}{year}\frac{MeV}{1.60\times10^{-13}J}=1.894\times10^{23}MeV$$
In the book, it is stated that the reaction in equation 31–8b starts with 2 deuterons, and releases 3.27 MeV of energy. Change this energy to an amount of deuterium.
$$\frac{ 1.894\times10^{23}MeV}{y}\frac{2fission}{3.27\;MeV}\frac{2.014g}{6.02\times10^{23}atoms}\approx 0.39g$$
