Answer
Please see the work below.
Work Step by Step
(a) We know that
$H=kA\frac{\Delta T}{\Delta x}$
We plug in the known values to obtain:
$H=(0.11)(1)(\frac{20}{0.02})=110W$
(b) As $H=kA\frac{\Delta T}{\Delta x}$
We plug in the known values to obtain:
$H=(0.029)(1)(\frac{20}{0.02})=29W$
