Cross sectional area \(A =\pi r ^{2}=10^{-4} m ^{2}\) \(j =\frac{ i }{ A }=\left(\frac{ V }{ R }\right) \cdot \frac{1}{ A }=\frac{ E \ell}{ RA } \quad R =\frac{\rho \ell}{ A }\) \(j =\frac{10 \times 10}{10 \times 10^{-4}}=10^{5} A / m ^{2}\)

\(J=\sigma E \Rightarrow \frac{ E }{\rho}=\frac{ E \ell}{ RA }=\frac{10 \times 10 \times \pi}{10 \times 10^{-4} \times \pi}\)

\(\Rightarrow 10^{5} A / m ^{2}\)