Answer
$v = 0.526 c$
Work Step by Step
The proton was accelerated from rest by a potential difference of 165 MV, so it has a kinetic energy of 165 MeV. Use equation 26–5b.
$$KE=165MeV=(\gamma-1)mc^2=(\gamma-1)(938.3MeV)$$
$$\frac{1}{\sqrt{1-v^2/c^2}}-1=\frac{165MeV}{938.3MeV}$$
$$v=c\sqrt{1-\frac{1}{(1+165/938.3)^2}}=0.526c$$
