Let conductivity of steel $K_{\text {steel }}=k$ then from question
Conductivity of copper $K_{\text {copper }}=9 k$
$$
\begin{aligned}
& \theta_{\text {copper }}=100^{\circ} \mathrm{C} \\
& \theta_{\text {steel }}=0^{\circ} \mathrm{C} \\
& l_{\text {steel }}=l_{\text {copper }}=\frac{L}{2}
\end{aligned}
$$
From formula temperature of junction;
$$
\begin{aligned}
\theta & =\frac{K_{\text {copper }} \theta_{\text {copper }} l_{\text {steel }}+K_{\text {steel }} \theta_{\text {steel }} l_{\text {copper }}}{K_{\text {copper }} l_{\text {steel }}+K_{\text {steel }} l_{\text {copper }}} \\
& =\frac{9 k \times 100 \times \frac{L}{2}+k \times 0 \times \frac{L}{2}}{9 k \times \frac{L}{2}+k \times \frac{L}{2}} \\
& =\frac{\frac{900}{2} k L}{\frac{10 k L}{2}}=90^{\circ} \mathrm{C}
\end{aligned}
$$