Answer
The least center-to-center distance the alpha particle will be from the target nucleus is $~~8.8\times 10^{-14}~m$
Work Step by Step
We can assume that the initial electric potential energy of the system is zero.
The alpha particle can reach a distance such that the electric potential energy of the system is equal to the initial kinetic energy of the system.
We can find the distance $r$:
$K = U$
$K = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{r}$
$r = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{K}$
$r = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~\frac{(92)(1.6\times 10^{-19}~C)(2)(1.6\times 10^{-19}~C)}{0.48\times 10^{-12}~J}$
$r = 8.8\times 10^{-14}~m$
The least center-to-center distance the alpha particle will be from the target nucleus is $~~8.8\times 10^{-14}~m$.