Introduction:
In this problem, we are given two concentric conducting spheres with radii 'aR' and '3R', where 'a' is a constant. The potential at the inner sphere is given as 2V, and the potential at the outer sphere is given as V. We need to find the potential at a distance r = 3R/2 from the center of the spheres.

Explanation:
To solve this problem, we can use the concept of equipotential surfaces and the formula for potential due to a uniformly charged sphere.
1. Equipotential Surfaces:
Equipotential surfaces are imaginary surfaces where the potential at every point is the same. In the case of concentric spheres, the equipotential surfaces are spheres with the same center but different radii. The potential decreases as we move from the inner sphere to the outer sphere.

2. Potential due to a Uniformly Charged Sphere:
The potential at a point outside a uniformly charged sphere is given by the formula:
V = k*(Q/R)
where:
- V is the potential at the point
- k is the electrostatic constant
- Q is the charge on the sphere
- R is the radius of the sphere

3. Potential at r = 3R/2:
Since the two spheres are concentric, the potential at a point between them will be the same as the potential on the surface of the inner sphere. Therefore, the potential at r = 3R/2 will be the same as the potential on the surface of the sphere with radius 'aR'.

Using the formula for potential due to a uniformly charged sphere, we can write:
2V = k*(Q_inner/aR) (1)
V = k*(Q_outer/(3R)) (2)

4. Relation between Charges:
Since the spheres are conductors, the charges on them will distribute uniformly on their surfaces. Therefore, we can write:
Q_inner = k_inner * (4π*(aR)^2) (3)
Q_outer = k_outer * (4π*(3R)^2) (4)

5. Relation between Potentials:
Substituting equations (3) and (4) into equations (1) and (2), we get:
2V = k_inner * (4π*(aR)^2)/(aR) (5)
V = k_outer * (4π*(3R)^2)/(3R) (6)

6. Simplifying Equations:
Simplifying equations (5) and (6), we get:
2V = 4π*k_inner*(aR)
V = 4π*k_outer*(3R)

7. Relation between Constants:
Since k_inner and k_outer are constants, we can write:
k_inner*(aR) = k_outer*(3R)

8. Calculating Potential at r = 3R/2:
Substituting the value of k_inner*(aR) from equation (8) into equation (5), we get:
2V = 4π*k_outer*(3R)*(3R/2)
2V = 6π*k_outer*(3R)^2