Answer
$k=0.1\frac{W}{m^{\circ}C}$
Work Step by Step
Knowing that the heat transfer in this case is by conduction:
$Q_{cond}=\frac{kA\Delta T}{L}$
Solving for $k$:
$k=\frac{Q_{cond}L}{A\Delta T}=\frac{Q_{cond}}{A}*\frac{L}{\Delta T}=500\frac{W}{m^2}*(\frac{0.02m}{100^{\circ}C-0^{\circ}C})=0.1\frac{W}{m^{\circ}C}$
