Like Eternal Code said, using a cylinder inside the original problem cylinder is the right approach. If you use Gauss' Law, you should find that the electric field inside the infinitely long, uniformly charged cylinder is

E=ρr/(2ε)

Now, to calculate the potential difference between the surface and axis of the cylinder,

deltaV=-∫ρr/(2ε)dr from 0 to R

This gives the potential difference between the surface and axis of the cylinder as being

deltaV=(-ρ(R^2))/(4ε0)

Like Eternal Code said, using a cylinder inside the original problem cylinder is the right approach. If you use Gauss' Law, you should find that the electric field inside the infinitely long, uniformly charged cylinder is

$$E=\frac{ρr}{(2ε_0)} $$

Now, to calculate the potential difference between the surface and axis of the cylinder,

$${\Delta V}=-\int_0^R \frac{ρr}{(2ε_0)}dr$$

This gives the potential difference between the surface and axis of the cylinder as being

$${\Delta V}=\frac{-ρ(R^2)}{4ε_0}$$