Answer
$4\times 10^7\;\rm m/s$
Work Step by Step
Let's assume that the system [the proton+ iron nucleus] is isolated. So the energy is conserved.
$$E_i=E_f$$
$$K_i+U_i=K_f+U_f$$
We know that the iron nucleus remains constant when the proton is fired and when the proton is on its surface since it reaches it at a speed of 0 m/s.
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 $r=r_{nucleus}$.
Let's assume that the proton is dimensionless as a point charge.
$$\frac{1}{2}m_pv_i^2+0= \dfrac{k(e)(26e)}{r_{nucleus}}$$
$$ v_i^2 = \dfrac{2k (26e^2)}{m_pr_{nucleus}}$$
$$ v_i = \sqrt{\dfrac{2k (26e^2)}{m_pr_{nucleus}}}$$
Plug the known;
$$ v_i = \sqrt{\dfrac{2(9\times 10^9) (26)(1.6\times 10^{-19})^2}{(1.67\times 10^{-27}) (4.5\times 10^{-15})}}$$
$$v_i=\color{red}{\bf 4\times 10^7}\;\rm m/s$$