$A,$ $\mathrm{B}$ and $\mathrm{C}$ respectively. Then

$\mathrm{R}_{\mathrm{A}}=\frac{2 l}{\mathrm{KA}} ; \mathrm{R}_{\mathrm{B}}=\frac{l}{2 \mathrm{K} \cdot 2 \mathrm{A}}=\frac{l}{4 \mathrm{KA}}$

and $\mathrm{R}_{\mathrm{C}}=\frac{l}{4 \mathrm{K} \cdot 2 \mathrm{A}}=\frac{l}{8 \mathrm{KA}}$

since thermal current

$\mathrm{H}=\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\theta_{1}-\theta_{2}}{l / \mathrm{KA}}=\frac{\theta_{1}-\theta_{2}}{\mathrm{R}}$

or $\quad \mathrm{RH}=\theta_{1}-\theta_{2}$

$\therefore \quad 100-\theta=\mathrm{R}_{\mathrm{A}} \mathrm{H}_{\mathrm{A}}$     $...(i)$

$\theta-50=\mathrm{R}_{\mathrm{B}} \mathrm{H}_{\mathrm{B}}$        $...(ii)$

$\theta-0=\mathrm{R}_{\mathrm{c}} \mathrm{H}_{\mathrm{c}}=\mathrm{R}_{\mathrm{C}}\left(\mathrm{H}_{\mathrm{A}}-\mathrm{H}_{\mathrm{B}}\right)$    $...(iii)$

On substituting values of $\mathrm{R}_{\mathrm{A}}, \mathrm{R}_{\mathrm{B}}$ and $\mathrm{R}_{\mathrm{C}},$ we get

$100-\theta=\frac{2 l}{\mathrm{KA}} \cdot \mathrm{H}_{\mathrm{A}} \Rightarrow \mathrm{H}_{\mathrm{A}}=\frac{\mathrm{K} \mathrm{A}}{2 l}(100-\theta)$

$\theta-50=\frac{l}{4 \mathrm{KA}} \mathrm{H}_{\mathrm{B}} \Rightarrow \mathrm{H}_{\mathrm{B}}=\frac{4 \mathrm{KA}}{l}(\theta-50)$

$\theta=\frac{l}{8 \mathrm{KA}}\left(\mathrm{H}_{\mathrm{A}}-\mathrm{H}_{\mathrm{B}}\right)$

$=\frac{1}{8 \mathrm{KA}}\left[\frac{\mathrm{KA}}{2 l}(100-\theta)-\frac{4 \mathrm{KA}}{l}(\theta-50)\right]$

$\theta=20^{\circ} \mathrm{C}$