As the temperature of rod varies from $\theta_1$ to $\theta_2$ from one end to another, so mean temperature of root at $C$ point.
Let us consider the diagram.
$$
\theta=\left(\frac{\theta_1+\theta_2}{2}\right)
$$
$\begin{array}{rrr}A & C & B \\ \theta_1 \bigcirc & \theta & \Theta^\theta\end{array}$

Let temperature varies linearly in the rod from its one end to other end. Let $\theta$ be the temperature of the mid-point $(C)$ of the rod. As steady state, so rate of flow of heat from $A$ to $C$ and $C$ to $B$ are equal. $\left(\theta_1>\theta>\theta_2\right)$
Then rate of flow of heat,
$$
\left(\frac{d Q}{d t}\right)=\frac{K A\left(\theta_1-\theta\right)}{\left(L_0 / 2\right)}=\frac{K A\left(\theta-\theta_2\right)}{\left(L_0 / 2\right)}
$$
where, $K$ is coefficient of thermal conductivity of the rod.
$$
\Rightarrow \theta_1-\theta=\theta-\theta_2 \Rightarrow \theta=\frac{\theta_1+\theta_2}{2}
$$
We know that,
$$
L=L_0(1+\alpha \theta)=L_0\left[1+\alpha\left(\frac{\theta_1+\theta_2}{2}\right)\right]
$$