Answer
See explanation.
Work Step by Step
a. Use the atomic masses given in the problem to calculate the mass defect of 0.1860u, the total masses on the left minus the total masses on the right. This equals an energy of 173.3 MeV.
b. In a gram of uranium-235, there are $\frac{0.001kg}{235m_p}=2.55\times10^{21}$ nuclei. The energy release per gram is $(173.3MeV/nucleus)(2.55\times10^{21}nuclei/g)=4.42\times10^{23}MeV/g$.
