Assume a gold nucleus is a sphere of radius 6.9 fermi in which...

The maximum possible separation between two protons in the nucleus is its diameter.

$r = 2R = 2 \times 6.9 \text{ fm} = 13.8 \times 10^{-15} \text{ m}$

The electrostatic repulsive force between the protons is given by Coulomb's law:

$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2}$

Substituting the values:

$F = \dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(13.8 \times 10^{-15})^2}$

$F = \dfrac{9 \times 10^9 \times 2.56 \times 10^{-38}}{190.44 \times 10^{-30}}$

$F \approx 1.2 \text{ N}$

The protons do not fly apart because the attractive strong nuclear force between nucleons is much stronger than the electrostatic repulsion at these distances.