In steady state the amount of heat flowing from one face to the other face in time $t$ is given by $Q=\frac{K A\left(\theta_1-\theta_2\right) t}{l}$ where $K$ is coefficient of thermal conductivity of material of rod $\Rightarrow \quad \frac{\mathrm{Q}}{\mathrm{t}} \propto \frac{\mathrm{A}}{\mathrm{l}} \propto \frac{\mathrm{r}^2}{\mathrm{l}}$
As the metallic rod is melted and the material is formed into a rod of half the radius
$$
\begin{aligned}
\mathrm{V}_1 & =\mathrm{V}_2 \\
\Rightarrow \quad \pi \mathrm{r}_1^2 \mathrm{l}_1 & =\pi \mathrm{r}_2^2 \mathrm{l}_2 \\
\quad \quad \quad \mathrm{l}_1 & =\frac{\mathrm{l}_2}{4}
\end{aligned}
$$
Now, from Eqs. (i) and (ii)
$$
\begin{aligned}
\Rightarrow \quad & \frac{Q_1}{Q_2}=\frac{r_1^2}{l_1} \times \frac{l_2}{r_2^2}=\frac{r_1^2}{l_1} \times \frac{4 l_1}{\left(r_1 / 2\right)^2} \\
\Rightarrow \quad Q_1 & =16 Q_2
\end{aligned}
$$