Three rods of identical cross-sectional area and made from the same...

Let ${T}_{0}$ be the temperature of point $C$ and $x$ be the length of rod $AB$ or $BC$.

Then, $CA = \sqrt{x^{2} + x^{2}} = \sqrt{2}x$

At steady state, the rate of heat flowing from $B$ to $C =$ rate of heat flowing from $C$ to $A$.

So, $\dfrac{KA\left(\sqrt{2}T - {T}_{0}\right)}{x} = \dfrac{KA\left({T}_{0} - T\right)}{\sqrt{2}x}$

$\sqrt{2}\left(\sqrt{2}T - {T}_{0}\right) = {T}_{0} - T$

By solving, ${T}_{0} = \dfrac{3T}{( \sqrt{2} + 1 )}$