Answer
a) $43500\frac{W}{m^2*^{\circ}C}$
Work Step by Step
First we calculate the lateral area of the electric resistance:
$A_{l}=\pi DL=\pi*0.002m*0.5m=0.003142m^2$
Knowing that the heat transfer in this case is by convection and that the boiling temperature of water at $1atm$ is $100^{\circ}C$
$Q_{conv}=hA\Delta T$
$h=\frac{Q_{conv}}{A\Delta T}=\frac{4100W}{0.003142m^2*(130^{\circ}C-100^{\circ}C}=43496.71\frac{W}{m^2*^{\circ}C}$
