Answer
$T=833.14^{\circ}C$
Work Step by Step
Knowing that the heat transfer in this case is by convection and radiation:
$Q_{conv}=hA\Delta T=20\frac{W}{m^2*^{\circ}C}*0.02m^2*(T_{s}-23^{\circ}C)=0.4(T_{s}-23^{\circ}C)W$
$Q_{rad}=\varepsilon \sigma A\Delta T^4=0.4*5.67*10^-8\frac{W}{m^2*K^4}*0.02m^2*[(T_{s}+273.15)^4-(23+273.15)^4]K^4=(4.536*10^-10*[(T_{s}+273.15)^4-(296.15)^4]K^4)W$
Then:
$Q_{t}=Q_{conv}+Q_{rad}$
$1000W=0.4(T_{s}-23^{\circ}C)W+(4.536*10^-10*[(T_{s}+273.15)^4-(296.15)^4]K^4)W$
Solving for $T$:
$T=833.14^{\circ}C$
