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
$${\Delta V}=\frac{-ρ(R^2)}{4ε_0}$$
