Answer
$8\times10^{12}J$.
Work Step by Step
Find the number of U-235 nuclei.
$$(2.0\;kg) \frac{6.02\times10^{23}U_{238}}{0.238kg}(0.05\frac{U_{235}}{U_{238}})=2.53\times10^{23}$$
We are given the information in problems 18-20 that 200 MeV of energy is released per fission. Multiply 200 MeV by the number of fissions, which is the number of U-235 atoms.
$$E=(200MeV)(2.53\times10^{23}) \frac{1.602\times10^{-13}J}{1MeV}\approx 8\times10^{12}J$$