If K _ 1 and K _ 2 are the thermal conductivities L _ 1 and L _ 2...

At steady state, the heat conducted by both the rods will be the same.

The rate of heat conduction is given by, $\dfrac{d Q}{d t} = \dfrac{KA}{l} ∆ T$. Therefore, if the temperature of the junction is $T$,

$\dfrac{{K}_{1}{A}_{1}}{{l}_{1}}\left({T}_{1} - T\right) = \dfrac{{K}_{2}{A}_{2}}{{l}_{2}}\left(T - {T}_{2}\right)$

$⇒ \dfrac{450 - T}{T - 0} = \dfrac{{K}_{2}{A}_{2}{l}_{1}}{{K}_{1}{A}_{1}{l}_{2}} = 9 \times \dfrac{1}{2} \times 2$

$⇒ 450 - T = 9T ⇒ T = 45 ° C$