= Solution
\b[The average moderator temperature is $725\,{}^\circ\mathrm C$.] First the mean coolant temperature in the <lumped thermal model> is
$$
\overline T_C=\frac{600+700}{2}=650\,{}^\circ\mathrm C.
$$
At a <steady state>, the same thermal <power> crosses both <thermal resistances>. If $R_1=R_2$, then
$$
\overline T_F-\overline T_M
=\overline T_M-\overline T_C.
$$
Thus
$$
\boxed{\overline T_M=\frac{800+650}{2}=725\,{}^\circ\mathrm C.}
$$
In particular $R_1=R_2=(800-650)/(2P)=3.17797\times10^{-7}\ \mathrm{K\,W^{-1}}$ at the specified state.
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