Correct option is (c) 0.5 × 10^-1 V/m
The explanation: Resistance of a wire = \(\frac {Ꝭl}{A}\), where Ꝭ is the specific resistance of the material of the wire.
Potential gradient = \(\frac {V}{l} \rightarrow \frac {IR}{l}\) → \(\frac {I(\frac {Ꝭl}{A})}{l}\)
= \(\frac {Ꝭl}{A}\)
= 0.5 × \(\frac {10^{-9}}{10^{-8}}\)
= 0.5 × 10^-1 V/m
Therefore, the potential gradient is equal to 0.5 × 10^-1 V/m.