= Solution
At a <steady state>, all three accumulation terms vanish. The fuel balance gives $Q_{FM}=P$, the moderator balance then gives $Q_{MC}=P$, and the coolant balance equates that input to its advective output. Therefore
$$
\boxed{
\begin{aligned}
\overline T_C&=T_i+\frac{P}{2W_Cc_C},\\
\overline T_M&=T_i+P\left(R_2+\frac1{2W_Cc_C}\right),\\
\overline T_F&=T_i+P\left(R_1+R_2+\frac1{2W_Cc_C}\right).
\end{aligned}}
$$
Each successive region adds the temperature drop needed to carry the same thermal <power> through its <thermal resistance>. Summing the original balances cancels both internal transfers, providing the independent total energy balance $P=W_Cc_C(T_o-T_i)$.
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