b
Let \( j\) be the current density.

Then \(j \times 2 \pi r^{2}=I \Rightarrow j=\frac{I}{2 \pi r^{2}} \therefore E=\rho j=\frac{\rho I}{2 \pi r^{2}}\)

Now, \(\Delta V_{BC}^\prime  =  - \int\limits_{a + b}^a {\vec E} .\overline {dr}  =  - \int\limits_{a + b}^a {\frac{{\rho I}}{{2\pi {r^2}}}dr} \)

\(=-\frac{\rho I}{2 \pi}\left[-\frac{1}{r}\right]_{a+b}^{a}=\frac{\rho I}{2 \pi a}-\frac{\rho I}{2 \pi(a+b)}\)

On applying superposition as mentioned we get

\(\Delta V_{B C}=2 \times \Delta V_{B C}^{\prime}=\frac{\rho I}{\pi a}-\frac{\rho I}{\pi(a+b)}\)