Two hollow conducting concentric spheres of radii R and r have positive charges with equal surface charge densities (σ). What is the electric potential at their common center? (r>R)

The inner conducting sphere has radius $R_1 = 2$ m and charge $Q_1 = Q$.
The outer conducting sphere has radius $R_2 = 6$ m and charge $Q_2 = -Q$.
The electric potential $V$ at a distance $r$ from the center of a conducting sphere with charge $Q'$ and radius $R'$ is given by:

$$V(r) = \begin{cases} \frac{kQ'}{R'} & \text{for } r \le R' \\ \frac{kQ'}{r} & \text{for } r > R' \end{cases}$$

where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

The potential on the inner sphere ($r=R_1$) is due to its own charge $Q_1$ and the charge $Q_2$ on the outer sphere.
The point $r=R_1$ is on the surface of the inner sphere and inside the outer sphere.
Potential due to $Q_1$: $V_{1,Q_1} = \frac{kQ_1}{R_1} = \frac{kQ}{2}$
Potential due to $Q_2$: Since $r=R_1 < R_2$, the potential due to the outer sphere at this point is the same as its surface potential. $V_{1,Q_2} = \frac{kQ_2}{R_2} = \frac{k(-Q)}{6} = -\frac{kQ}{6}$
The total potential on the inner sphere is:

$$V_1 = V_{1,Q_1} + V_{1,Q_2} = \frac{kQ}{2} - \frac{kQ}{6} = kQ \left( \frac{1}{2} - \frac{1}{6} \right) = kQ \left( \frac{3-1}{6} \right) = kQ \left( \frac{2}{6} \right) = \frac{kQ}{3}$$

The potential on the outer sphere ($r=R_2$) is due to the charge $Q_1$ on the inner sphere and its own charge $Q_2$.
The point $r=R_2$ is outside the inner sphere and on the surface of the outer sphere.
Potential due to $Q_1$: Since $r=R_2 > R_1$, $V_{2,Q_1} = \frac{kQ_1}{R_2} = \frac{kQ}{6}$
Potential due to $Q_2$: $V_{2,Q_2} = \frac{kQ_2}{R_2} = \frac{k(-Q)}{6} = -\frac{kQ}{6}$
The total potential on the outer sphere is:

$$V_2 = V_{2,Q_1} + V_{2,Q_2} = \frac{kQ}{6} - \frac{kQ}{6} = 0$$

The potential difference between the spheres is $\Delta V = V_1 - V_2$.

$$\Delta V = \frac{kQ}{3} - 0 = \frac{kQ}{3}$$

The calculated potential difference is $\frac{kQ}{3}$, which matches option (2).

The final answer is $\text{(2) } \frac{kQ}{3}$.