Answer
$Q_{cond}=124.8W$
Work Step by Step
Termal conductivity of air is:
$k=0.026\frac{W}{m^{\circ}C}$
And the area of the glass:
$A=2m*2m=4m^2$
Knowing that the heat transfer in this case is by conduction:
$Q_{cond}=\frac{kA\Delta T}{L}=\frac{0.026\frac{W}{m^{\circ}C}*4m^2*(18^{\circ}C-6^{\circ}C)}{0.01m}=124.8W$
