Answer
$$84.3^{\circ} \mathrm{C}
$$
Work Step by Step
We note that $ T_{H}=100^{\circ} \mathrm{C}, T_{C}=0^{\circ} \mathrm{C},$
copper-aluminum temperature junction by $T_{1}$.
and also we note that aluminum-brass junction by $T_{2}$.
Then,
$$
P_{\text {cond }}=\frac{k_{c} A}{L}\left(T_{H}-T_{1}\right)=\frac{k_{a} A}{L}\left(T_{1}-T_{2}\right)=\frac{k_{b} A}{L}\left(T_{2}-T_{c}\right)
$$
We solve for $T_{1}$ and $T_{2}$ to get
$$
T_{1}=T_{H}+\frac{T_{C}-T_{H}}{1+k_{c}\left(k_{a}+k_{b}\right) / k_{a} k_{b}}$$
$$=100^{\circ} \mathrm{C}+\frac{0.00^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}}{1+401(235+109) /[(235)]}$$
$$=84.3^{\circ} \mathrm{C}
$$
