Answer
${\bf 1.77\times 10^7}\;\rm V$
Work Step by Step
To solve this problem, we can use the principle of conservation of energy. Initially, the alpha particle has only kinetic energy, and as it moves closer to the $\rm ^{238}U$ 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 $Q=Z_{\rm ^238U}e=92e$ is the charge of the nucleus, $2e$ is the charge of the alpha particle, $r_i=\infty$, and $r_f=r_{\rm ^238U}=7.5\;\rm fm$, $K_f=0$
$$0+K_i=\dfrac{k(2e)(92e)}{r_{\rm ^238U}}+0$$
$$ K_i=\dfrac{184ke^2}{r_{\rm ^238U}} $$
Recalling that $\Delta K=q\Delta V$, so
$$ (2e)\Delta V=\dfrac{184ke^2}{r_{\rm ^238U}} $$
Thus,
$$ \Delta V=\dfrac{184ke}{2 r_{\rm ^238U}} $$
Plug the known;
$$ \Delta V=\dfrac{184(9\times 10^9)(1.6\times 10^{-19})}{2(7.5\times 10^{-15})} $$
$$\Delta V=\color{red}{\bf 1.77\times 10^7}\;\rm V$$
