c
(c) $\frac{Q}{t} = \frac{{KA\Delta \theta }}{l}$
$⇒$ $\frac{{mL}}{t}$ $ = \frac{{K(\pi {r^2})\Delta \theta }}{l}$
$⇒$ Rate of melting of ice $\left( {\frac{m}{t}} \right) \propto \frac{{K{r^2}}}{l}$
Since for second rod $K$ becomes $\frac{1}{4}th$ $r$ becomes double and length becomes half, so rate of melting will be twice i.e. ${\left( {\frac{m}{t}} \right)_2} = 2\,{\left( {\frac{m}{t}} \right)_1} = 2 \times 0.1 = 0.2\,gm/sec.$