Answer
(a) $0.21KW$
(b) $0.42KW$
Work Step by Step
(a) We know that
$\frac{Q}{t}=\frac{KA\Delta T}{L}$
We plug in the known values to obtain:
$\frac{Q}{t}=\frac{(0.60W/m-K)(1.40m^2)(3K)}{1.20\times 10^{-2}m}$
$\implies \frac{Q}{t}=210W=0.21KW$
(b) We know that the rate of transfer of the heat is doubled, that is, $420W=0.42KW$ when the temperature difference is doubled.