Answer
a). $0.13ohm$
b). $0.038ohm$
Work Step by Step
a). Resistivity at $380^{\circ}C$ is
$=p(1+alpha\times\Delta T)=10^{-7}(1+6.51\times10^{-3}\times380)=3.474\times10^{-7}ohm-m$
So, resistance at $380^{\circ}C$ = $p\frac{L}{A}=\frac{3.474\times10^{-7}\times0.75}{2\times10^{-6}}=0.13ohm$
b). Resistance of the coil when stove is off is equal to that when in room temperature, which is $=p\frac{L}{A}=\frac{10^{-7}\times0.75}{2\times10^{-6}}=0.038ohm$