Answer
See the detailed answer below.
Work Step by Step
We need to find the proton's kinetic energy to find its speed. So we will solve part b first.
$$\color{blue}{\bf [b]}$$
To solve this problem, we can use the principle of conservation of energy. Initially, the proton has only kinetic energy, and as it moves closer to the $\rm ^{16}O$ nucleus, some of this kinetic energy is converted into electric potential energy.
$$U_i+K_i=U_f+K_f$$
$$\dfrac{kQq}{r_i}+K_i=\dfrac{kQq}{r_f}+K_f$$
where $Q=Z_{\rm ^{16}O}e=8e$ is the charge of the nucleus, $e$ is the charge of the proton, $r_i=\infty$, and $r_f=1\;{\rm fm}+r_{\rm ^{16}O}=4\;\rm fm$, $K_f=0$
$$0+K_i=\dfrac{k( e)(8e)}{r_f}+0$$
$$ K_i=\dfrac{8ke^2}{r_f} $$
Plug the known;
$$ K_i=\dfrac{8(9\times 10^9)(1.6\times 10^{-19})^2}{ (4\times 10^{-15})} $$
$$ K_i=\bf 4.608 \times 10^{-13}\;\rm J=\color{red}{\bf 2.88}\;\rm Mev$$
$$\color{blue}{\bf [a]}$$
Recalling that
$$K_i=\frac{1}{2}mv_i^2$$
So,
$$v_i=\sqrt{\dfrac{2K_i}{m}}=\sqrt{\dfrac{2 (4.608 \times 10^{-13})}{(1.67\times 10^{-27})}}$$
$$v_i=\color{red}{\bf2.35\times 10^7}\;\rm m/s$$
