Let $K_A$ and $K_B$ be the thermal conductivity of material $A$ and $B$, respectively.
Let $T_1$ be the temperature of $A$ and $T_2$ be the temperature of $B$.
At steady state, rate of flow of heat is given by,
$$
\begin{aligned}
\frac{\Delta Q}{\Delta t} & =\frac{\Delta Q_A}{\Delta t}=\frac{\Delta Q_B}{\Delta t} \\
\Rightarrow \frac{K^{\prime} A\left(T_2-T_1\right)}{l_A+l_B} & =\frac{K_A A\left(T-T_1\right)}{l_A} \\
& =\frac{K_B A\left(T_2-T\right)}{l_B}
\end{aligned}
$$
where $K$ is thermal conductivity of both material and $A$ and $l$ are area and length of the material respectively.
Using Eq. (i), we get
$$
\begin{aligned}
& \frac{K_A A}{l_A}\left(T-T_1\right)=\frac{K_B A\left(T_2-T\right)}{l_B} \\
& \Rightarrow \quad 2 K_B\left(T-T_1\right)=K_B\left(T_2-T\right) \\
& \left(\because l_A=l_B \text { and } K_A=2 K_B\right) \\
& \Rightarrow \quad 2 T-2 T_1=T_2-T \\
& \Rightarrow \quad 3 T=T_2+2 T_1 \Rightarrow T=\frac{2 T_1+T_2}{3} \\
&
\end{aligned}
$$
Again using Eq. (i), we get
$$
\begin{aligned}
& \frac{K^{\prime} A\left(T_2-T_1\right)}{l_A+l_B}=\frac{K_A A\left(T-T_1\right)}{l_A} \\
\Rightarrow \quad \frac{K^{\prime}\left(T_2-T_1\right)}{2 l_A} & =\frac{K_A\left(\frac{2 T_1+T_2}{3}-T_1\right)}{l_A} \\
\Rightarrow \quad & \frac{K^{\prime}\left(T_2-T_1\right)}{2}=\frac{K_A\left(T_2-T_1\right)}{3}
\end{aligned}
$$
$$
\Rightarrow \quad K^{\prime}=\frac{2}{3} K_A=\frac{2}{3} \times 2 K_B \Rightarrow K^{\prime}=\frac{4}{3} K_B
$$
Using Eq. (i) to evaluate $\left(T_2-T\right)$,
$$
\begin{aligned}
& \frac{K^{\prime} A\left(T_2-T_1\right)}{l_A+l_B}=\frac{K_B A\left(T_2-T\right)}{l_B} \\
\Rightarrow & \frac{K^{\prime} A\left(T_2-T_1\right)}{2 l_A}=\frac{K_B A\left(T_2-T\right)}{l_A} \\
\Rightarrow & \left(\frac{4}{3} K_B\right) \frac{A\left(T_2-T_1\right)}{2}=K_B A\left(T_2-T\right) \\
\Rightarrow & \frac{2}{3}\left(T_2-T_1\right)=T_2-T \\
\Rightarrow & T_2-T=\frac{2}{3}\left(T_2-T_1\right)=\frac{2}{3} \times 24^{\circ}=16^{\circ} \mathrm{C}
\end{aligned}
$$