Answer
$E = 3.07\times 10^{21}~N/C$
Work Step by Step
The electric field at the surface of a sphere with the charge uniformly spread throughout the sphere is the same as the electric field at the surface of a sphere if all the charge is concentrated at the point at the center of the sphere.
We can find the magnitude of the electric field at the surface of the nucleus:
$E = \frac{q}{4\pi~\epsilon_0~r^2}$
$E = \frac{(94)(1.6\times 10^{-19}~C)}{(4\pi)(8.854\times 10^{-12}~C^2/N~m^2)~(6.64\times 10^{-15}~m)^2}$
$E = 3.07\times 10^{21}~N/C$
