c
Radius of wire $=\frac{10^{-2}}{\sqrt{\pi}}$
Cross sectional area $A =\pi r ^{2}=10^{-4} m ^{2}$ $j =\frac{ i }{ A }=\left(\frac{ V }{ R }\right) \cdot \frac{1}{ A }=\frac{ E \ell}{ RA } \quad R =\frac{\rho \ell}{ A }$ $j =\frac{10 \times 10}{10 \times 10^{-4}}=10^{5} A / m ^{2}$
$J=\sigma E \Rightarrow \frac{ E }{\rho}=\frac{ E \ell}{ RA }=\frac{10 \times 10 \times \pi}{10 \times 10^{-4} \times \pi}$
$\Rightarrow 10^{5} A / m ^{2}$