Answer
$4.0\times10^{-14}m$.
Work Step by Step
We are told to ignore the recoil of the nucleus. At the closest approach, the kinetic energy of the system is zero. The final potential energy of the two charges equals the kinetic energy of the incoming $\alpha$ particle.
$$KE_{\alpha}=\frac{1}{4\pi \epsilon_o}\frac{(2e)(79e)}{r_{min}}$$
$$ r_{min}=\frac{1}{4\pi \epsilon_o}\frac{(2e)(79e)}{ KE_{\alpha}}$$
$$ =(9.00\times10^9Nm^2/C^2)\frac{(2)(79)(1.60\times10^{-19}C)^2}{(4.8MeV)(1.60\times10^{-13}J/MeV)}$$
$$=4.74\times10^{-14}m$$
The distance to the surface is this center-to-center spacing minus the radius of the gold nucleus, $4.74\times10^{-14}m-7.0\times10^{-15}m=4.0\times10^{-14}m$.
