given total charge Q uniformly distributes in an atomic nucleus with radius R.
If you mean the external electric potential due to the atomic nucleus then I believe the answer is exactly the same as the electric field for a point charge at the nucleus centre. So if you are a distance $\displaystyle Z$ from the nucleus centre ($\displaystyle Z>R$) then the potential is
$\displaystyle \Phi_E = \frac{Q}{4\pi\epsilon_0 Z}$
If you need to show this what you need to do is perform a volume integration in spherical coords. Your density is $\displaystyle \rho=3Q / 4\pi R^3$, and the distance to the volume element is $\displaystyle \delta = (r^2+Z^2-2rZ\cos\phi)^{1/2}$. Then you want to evaluate
$\displaystyle
\Phi_E = \frac{\rho}{4\pi\epsilon_0}\int_0^R\int_0^\pi\int_ 0^{2\pi} \frac{r^2\sin\phi\,d\theta\,d\phi\,dr}{(r^2+Z^2-2rZ\cos\phi)^{1/2}}.$
If we assume $\displaystyle Z>r$ this integral is easy and gives the formula above.
I hope this is what you're after.