Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
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 $\rm ^{12}C$ 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 $Q=Z_{\rm ^{12}C}e=6e$ is the charge of the nucleus, $e$ is the charge of the proton, $r_i=\infty$, and $r_f= r_{\rm ^{12}C}=2.75\;\rm fm$, $K_f=3$ MeV.
$$0+K_i=\dfrac{k( e)(6e)}{r_f}+K_f$$
$$ K_i=\dfrac{6ke^2}{r_f} +K_f$$
$$ \frac{1}{2}mv_i^2=\dfrac{6ke^2}{r_f} +K_f$$
Hence,
$$v_i=\sqrt{\dfrac{\dfrac{6ke^2}{r_f} +K_f}{\frac{1}{2}m}}$$
Plug the known;
$$v_i=\sqrt{\dfrac{\dfrac{6(9\times 10^9)(1.6\times 10^{-19})^2}{ (2.75\times 10^{-15})}+(3\times 10^6\times 1.6\times 10^{-19})}{\frac{1}{2}(1.67\times 10^{-27})}}$$
$$v_i=\color{red}{\bf 3.43 \times 10^7}\;\rm m/s$$
$$\color{blue}{\bf [b]}$$
Now we need to find the potential difference the proton must be accelerated from rest to acquire this speed.
Recall that
$$K_i=\frac{1}{2}mv_i^2=e\Delta V$$
So,
$$\Delta V=\dfrac{\frac{1}{2}mv_i^2}{e}$$
Plug the known;
$$\Delta V=\dfrac{\frac{1}{2}(1.67\times 10^{-27})( 3.43 \times 10^7)^2}{1.6\times 10^{-19}}$$
$$\Delta V=\color{red}{\bf 6.14\times 10^6}\;\rm V$$
