Answer
$T_{outersurface}=105.564^{\circ}C$
Work Step by Step
First we calculate the are of the pan:
$A=\frac{\pi*(0.2m)^2}{4}=0.031416m^2$
Knowing that the heat transfer in this case is by conduction:
$Q_{cond}=\frac{kA\Delta T}{L}$
Solving for $\Delta T$:
$\Delta T= \frac{Q_{cond}L}{kA}=\frac{700W*0.006m}{237\frac{W}{m^{\circ}C}*0.031416m^2}=0.564^{\circ}C$
Taking $T_{1}$ as the temperature on the inner surface and $T_{2}$ as the temperature of the outer surface of the bottom of the pan:
$\Delta T=T_{2}-T_{1}$
$T_{2}=105^{\circ}C+0.564^{\circ}C=105.564^{\circ}C$
