Answer
$\rm 6.8\;fm$
Work Step by Step
Let's assume that the system [the proton+ mercury nucleus] is isolated. So the energy is conserved.
$$E_i=E_f$$
$$K_i+U_i=K_f+U_f$$
We know that the mercury nucleus remains constant when the proton is fired and when the proton is at its closest approach to the surface of the nucleus, its speed is zero.
Hence,
$$\frac{1}{2}m_pv_i^2+U_i=0+U_f$$
initially, the proton was very far away $r\rightarrow \infty$, and when the proton on the surface of the nucleus $r_f=d+r_{nucleus}$.
Let's assume that the proton is dimensionless as a point charge.
$$\frac{1}{2}m_pv_i^2+0= \dfrac{k(e)(80e)}{d+r_{nucleus}}$$
$$ d+r_{nucleus}= \dfrac{80ke^2}{\frac{1}{2}m_pv_i^2 }$$
$$ d= \dfrac{80ke^2}{\frac{1}{2}m_pv_i^2 }-r_{nucleus}$$
Plug the known;
$$ d= \dfrac{80(9\times 10^9)(1.6\times 10^{-19})^2}{\frac{1}{2}(1.67\times 10^{-27})(4\times 10^7)^2 }-(7\times 10^{-15})$$
$$d=\color{red}{\bf 6.8\times 10^{-15}}\;\rm m$$
