Answer
$ r_f = 1.64 \times 10^{-9} m $
Work Step by Step
$E = \Delta K + (-\Delta U)$ since
$E = \Delta \frac{1}{2} m v^2 - \Delta \frac{ke^2}{r}$
$E = (\frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2) - ( \Delta \frac{ke^2}{r_f} - \Delta \frac{ke^2}{r_i}) $
Since the proton is at rest initially, $U_f = 0$ and energy must be conserved, so $E = 0$
$0 = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2 - \frac{ke^2}{r_f}$
Now we solve for $r_f$ and $v_f = 2V_i$
$\frac{ke^2}{r_f} = \frac{1}{2} m(2v_i)^2 - \frac{1}{2} mv_i^2 $
$\frac{ke^2}{r_f} = \frac{1}{2} m (4v_i^2 - v_i^2) $
$\frac{ke^2}{r_f} = \frac{1}{2} m (3v_i^2 ) $
$ r_f = \frac{2ke^2}{3m v_i^2} $
$ r_f = \frac{2(8.99 \times 10^9 N . m^2 / C^2)(1.6 \times 10^{-19}C)^2}{3(9.11 \times 10^{-31} kg ) (3.2 \times 10^5 m/s)^2} $
$ r_f = 1.64 \times 10^{-9} m $
