Given, Mass of $U^{235} = 1 \text{ kg} = 1000 \text{ g}$. Number of moles in $1000 \text{ g}$ of $U^{235}$: $$ \begin{aligned} n &= \frac{1000}{235} \text{ mol} \end{aligned} $$ Number of atoms in $1000 \text{ g}$ of $U^{235}$: $$ \begin{aligned} N &= n \times N_A \\ &= \frac{1000}{235} \times 6.02 \times 10^{23} \text{ atoms} \end{aligned} $$ Energy released per fission reaction: $$ \begin{aligned} \Delta m &= (235.04 + 1.01) - (140.31 + 91.91 + 3 \times 1.01) \text{ amu} \\ &= 236.05 - 235.24 \text{ amu} \\ &= 0.81 \text{ amu} \end{aligned} $$ Energy released per atom: $$ \begin{aligned} E_{\text{atom}} &= 0.81 \times 932 \text{ MeV} \\ &= 754.92 \text{ MeV} \end{aligned} $$ Total energy released: $$ \begin{aligned} E_{\text{total}} &= N \times E_{\text{atom}} \\ &= \left( \frac{1000}{235} \times 6.02 \times 10^{23} \right) \times 754.92 \text{ MeV} \\ &= 4.264 \times 10^{26} \text{ MeV} \approx 4.26 \times 10^{26} \text{ MeV} \end{aligned} $$ Therefore, the total energy released is $4.26 \times 10^{26} \text{ MeV}$.