Answer
$82.25$ $kV$
Work Step by Step
Let, the proton would have to be accelerated through electric potential $V$ to acquire the speed $v=3.97\times 10^{6}$ $m/s$
Thus,
$eV=\frac{1}{2}mv^{2}$
or, $V=\frac{mv^{2}}{2e}$
where, $m$ is the mass of a proton and $e$ is the charge of a proton.
For, $v=3.97\times 10^{6}$ $m/s$
$V=\frac{1.67\times 10^{-27}\times(3.97\times 10^{6})^{2}}{2\times 1.6\times 10^{-19}}$ $V$
or, $V\approx 82252$ $V$
or, $V\approx82.25$ $kV$
Therefore, the proton would have to be accelerated through electric potential of $82.25$ $kV$
