Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities $$+\sigma$$, $$-\sigma$$ and $$+\sigma$$ respectively. The potential of shell B is:

We begin by recalling the expression for the electric potential produced by a uniformly charged thin spherical shell. A shell of total charge $$Q$$ and radius $$R$$ produces, at a point situated a distance $$r$$ from the common centre, a potential

$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}\quad\text{for}\;r\ge R,$$

while everywhere inside the shell (that is, for $$r<R$$) the potential is constant and equal to its value on the surface, viz.

$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}\quad\text{for}\;r\le R.$$

Each of the three metal shells A, B and C is concentric, so their individual contributions at any point simply add. The surface charge densities are given as $$+\sigma$$ on A (radius $$a$$), $$-\sigma$$ on B (radius $$b$$) and $$+\sigma$$ on C (radius $$c$$), with the order $$a<b<c$$.

First we evaluate the actual charges on the shells:

$$Q_A = \sigma\;(4\pi a^2),\qquad Q_B = (-\sigma)\;(4\pi b^2),\qquad Q_C = \sigma\;(4\pi c^2).$$

Now we calculate separately the potentials at the point on shell B, i.e. at radial distance $$r=b$$.

Contribution of shell A: the observation point lies outside shell A (because $$b>a$$). Using the outside-point formula we have

$$V_A(b)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_A}{b} =\dfrac{1}{4\pi\varepsilon_0}\dfrac{\sigma(4\pi a^2)}{b} =\dfrac{\sigma a^2}{\varepsilon_0 b}.$$

Contribution of shell B itself: for a point on a conducting shell the potential due to that shell equals the value just outside, obtained with the same outside-point formula but with $$r=b$$ (or, equivalently, with the inside constant value). Thus

$$V_B(b)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_B}{b} =\dfrac{1}{4\pi\varepsilon_0}\dfrac{-\sigma(4\pi b^2)}{b} =-\dfrac{\sigma b}{\varepsilon_0}.$$

Contribution of shell C: the observation point is inside shell C because $$b<c$$. Hence the potential is constant throughout the interior and equals its surface value, viz.

$$V_C(b)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_C}{c} =\dfrac{1}{4\pi\varepsilon_0}\dfrac{\sigma(4\pi c^2)}{c} =\dfrac{\sigma c}{\varepsilon_0}.$$

We now add these three contributions to obtain the total potential of shell B:

$$\begin{aligned} V_{\text{total}}(b) &=V_A(b)+V_B(b)+V_C(b)\\[4pt] &=\dfrac{\sigma a^2}{\varepsilon_0 b} -\dfrac{\sigma b}{\varepsilon_0} +\dfrac{\sigma c}{\varepsilon_0}\\[4pt] &=\dfrac{\sigma}{\varepsilon_0} \left(\dfrac{a^2}{b}-b+c\right). \end{aligned}$$

It is convenient to combine the first two terms over a common denominator $$b$$:

$$\dfrac{a^2}{b}-b=\dfrac{a^2-b^2}{b},$$

so the potential becomes

$$V_{\text{total}}(b)= \dfrac{\sigma}{\varepsilon_0} \left[\dfrac{a^2-b^2}{b}+c\right].$$

This expression matches exactly the option labelled

$$\dfrac{\sigma}{\varepsilon_0}\left[\dfrac{a^2-b^2}{b}+c\right],$$

which is Option C.

Hence, the correct answer is Option C.