= Solution
The bookkeeping reaction is
$$
n+{}^{235}_{92}\mathrm U\longrightarrow{}^{92}_{37}\mathrm{Rb}
+{}^{140}_{55}\mathrm{Cs}+4n.
$$
Both <mass number> and <electric charge> are conserved: $236=92+140+4$ and $92=37+55$. Express each atomic mass as the sum of its constituent rest masses minus its <nuclear binding energy>. The total constituent masses, including the atomic-electron masses, cancel between the two sides. Neglecting incident-neutron <kinetic energy>, the <Q value> is therefore
$$
\boxed{Q=B(37,92)+B(55,140)-B(92,235).}
$$
The relevant initial binding is that of uranium-235. Introducing uranium-236 as an intermediate state is possible only if the neutron-capture energy is also accounted for.
In the stated coefficient convention,
$$
B(Z,A)=15.56A-17.23A^{2/3}-0.697\frac{Z^2}{A^{1/3}}
-93.14\frac{(Z-A/2)^2}{A}+\delta_B.
$$
Uranium-235 has odd $A$, hence $\delta_B=0$. Both products are odd–odd: rubidium has $(Z,N)=(37,55)$ and caesium has $(55,85)$. Each therefore has $\delta_B=-12/\sqrt A$ MeV. Direct substitution gives
$$
\begin{aligned}
B(92,235)&=1786.751\ \mathrm{MeV},\\
B(37,92)&=785.759\ \mathrm{MeV},\\
B(55,140)&=1157.089\ \mathrm{MeV},\\
\boxed{Q}&\boxed{=156.098\ \mathrm{MeV}.}
\end{aligned}
$$
This is the estimate from the supplied smooth mass formula, rather than a value imported from measured isotope masses.
For $100$ MW recovered as the stated fission energy, the annual energy output is $10^8(365.25)(86400)$ J. Thus
$$
\boxed{N_{\rm fission/yr}=\frac{10^8(31557600)}{156.0976(1.602176634\times10^{-13})}
=1.262\times10^{26}.}
$$
Each event consumes one uranium-235 atom. With <Avogadro constant> $N_A=6.02214076\times10^{23}$ mol$^{-1}$ and molar mass approximated as $235$ g mol$^{-1}$,
$$
\boxed{m=\frac{235N_{\rm fission/yr}}{N_A}=4.92\times10^4\ \mathrm g\simeq49.2\ \mathrm{kg}.}
$$
This uses the stated power directly as recovered fission power; an unspecified conversion efficiency or additional heat-loss correction cannot be inferred from it.
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