Answer
$T = \frac{2\pi~m}{q~B}$
Work Step by Step
The force $F$ provides the centripetal force to keep the particle moving in a circle. We can find an expression for the speed;
$F = \frac{m~v^2}{r}$
$q~v~B = \frac{m~v^2}{r}$
$v = \frac{q~r~B}{m}$
We then find the period $T$ of the circular motion:
$T = \frac{distance}{speed}$
$T = \frac{2\pi~r}{v}$
$T = \frac{2\pi~r}{(q~r~B)/m}$
$T = \frac{2\pi~m}{q~B}$
