Answer
$9.9\times10^2$ kg of deuterium used in a year.
Work Step by Step
To produce 1000 MW of electrical power at a 33 percent efficiency, the actual power generated by the reaction of equation 31–8a is greater: 1000 MW/0.33 = 3030 MW.
Convert that to mass M of deuterium used in one year, using Eq. 31–8a.
$$M=(\frac{3.03\times10^9J}{s})(\frac{2\;d\;nuclei}{4.03MeV})(\frac{1MeV}{1.60\times10^{-13}J})(\frac{0.002kg\;d\;nuclei}{6.02\times10^{23}nuclei})(\frac{3.156\times10^7s}{yr})$$
$=9.9\times10^2$ kg of deuterium used in a year.
