Let the initial radius be $r$ and length be $l$. The volume of the rod is $V = \pi r^2 l$. The rate of heat conduction is given by $\frac{Q}{t} = \frac{KA\Delta T}{l} = \frac{K(\pi r^2)\Delta T}{l}$. When the rod is melted and formed into a new rod of radius $r' = \frac{r}{2}$, its volume remains constant. $V' = V \implies \pi (r')^2 l' = \pi r^2 l \implies \pi \left(\frac{r}{2}\right)^2 l' = \pi r^2 l \implies l' = 4l$. The new area of cross-section is $A' = \pi (r')^2 = \frac{\pi r^2}{4} = \frac{A}{4}$. The new amount of heat conducted in time $t$ is $Q' = \frac{KA'\Delta T}{l'} t = \frac{K(A/4)\Delta T}{4l} t = \frac{1}{16} \left(\frac{KA\Delta T}{l} t\right) = \frac{Q}{16}$.