In the book of physics it is Given that the Gravitational Potential Energy at the center of earth(Vc) = 3/2(Vs?

The gravitational potential energy of an object on the surface of the Earth is the integral of the acceleration of gravity from r = infinity to R:

∫ GM(earth)/r² dr

This quantity equals -GM(earth)/R.

If Earth is assumed to be a sphere of uniform density (it isn't, but for the sake of argument) the difference in energy between the surface and the center is also the integral of the acceleration of gravity from R to 0, but the acceleration of gravity follows a different curve.  Inside a sphere of uniform density the mass inside the radius decreases as the cube of the radius:  m = 4/3πρr³.  Since the acceleration is Gm/r², the acceleration is 4/3Gπρr.  If you work this out, the ΔE between r=0 and r=R is -½GM(earth)/R.

Add the two together and you get -3/2GM/R.

There is a real issue with assuming g = G rho 4/3 r^3/r^2 = G rho 4/3 r is the gravity acceleration within a sphere of uniform density. That issue is that as we bore towards the center of that sphere, more and more mass is above (back in the direction of the surface), which offsets the pull from the mass in the center of the sphere.

In other words, g is way overstated because it discounts the increasing pull of the mass above the interior point r < R. In fact, just prior to r = 0 at the center of the mass (e.g., the uniformly dense Earth), the pull of gravity would be equal in all directions outward towards the surface. In other words, there would be no net force to accelerate anything in the center as they would all cancel out.

My point is this... that relationship Vc = 3/2 Vs is highly flawed and unrealistic because it ignores the pull of gravity back outward toward the surface due to the mass lying from r --> R.