a
Let conductivity of steel \({K_{steel}{ = }}k\,then\)

From question

Conductivity of copper \({K_{copper}} = 9k\)

\({\theta _{copper}} = {100^ \circ }C\)

\({\theta _{steel}} = {0^ \circ }C\)

\({l _{steel}} = {l_{copper}} = \frac{L}{2}\)

From formula temperature of junction;

\(\theta  = \frac{{{K_{copper}}{\theta _{copper}}{l_{steel}} + {k_{steel}}{\theta _{steel}}{l_{copper}}}}{{{K_{copper}}{l_{steel}} + {K_{steel}}{l_{copper}}}}\)

\( = \frac{{9k \times 100 \times \frac{L}{2} + k \times 0 \times \frac{L}{2}}}{{9k \times \frac{L}{2} + k \times \frac{L}{2}}}\)

\( = \frac{{\frac{{900}}{2}}}{{\frac{{10kL}}{2}}} = {90^ \circ }C\)