Answer
$p = 3.59~MeV/c$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$\gamma = \frac{1}{\sqrt{1-0.990^2}}$
$\gamma = 7.0888$
We can find the momentum:
$p = \gamma mv$
$p = (\gamma)(E_0/c^2) (v)$
$p = (7.0888)(511~keV/c^2) (0.990~c)$
$p = 3.59~MeV/c$
