Answer
$Q_{cond}=1035W$
Work Step by Step
First we calculate the are of the wall:
$A=5m*6m=30m^2$
Knowing that the heat transfer in this case is by conduction:
$Q_{cond}=\frac{kA\Delta T}{L}=\frac{0.69\frac{W}{m^{\circ}C}*30m^2*(20^{\circ}C-5^{\circ}C)}{0.3m}=1035W$
