Answer
$T = 3.34\times 10^{-10}~s$
Work Step by Step
We can find $\gamma$:
$K = (\gamma-1)~mc^2 = 10.0~MeV$
$\gamma-1 = \frac{10.0~MeV}{mc^2}$
$\gamma = 1+\frac{10.0~MeV}{0.511~MeV}$
$\gamma = 20.57$
We can find the period:
$T = \frac{2\pi~\gamma ~m}{\vert q \vert B}$
$T = \frac{(2\pi)(20.57)(9.109\times 10^{-31}~kg)}{(1.6\times 10^{-19}~C)(2.20~T)}$
$T = 3.34\times 10^{-10}~s$
