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

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

As shown above $\quad E=\frac{\rho I}{2 \pi r^{2}}$