$(a)\;10^{-2} v/m \\(b)\;2 \times 10^{-2}\;v/m \\(c)\;10^{-3} \;v/m \\(d)\;3 \times 10^{-3}\;v/m $

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Potential gradient is $ \large\frac{V}{L}$ where V is potential across wire of length L.

$R= \large\frac{\rho L}{A}$

$\qquad= \large\frac{40 \times 10^{-8} \times L}{8 \times 10^{-6}}$

$\qquad= 5 \times 10^{-2}\;L$

$V= I \times R$

$\qquad= 5 \times 10^{-2} \times L \times 0.2$

$\qquad = 1 \times 10^{-2} \times L$

Potential gradient $=\large\frac{V}{L}$

$\qquad= 10^{-2}\;V/m$

Hence a is the correct answer.

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