In a cyclotron, the centripetal force is provided by the transverse magnetic field $B$, and the force on a particle travelling in a magnetic field (which causes it to curve) is equal to $B q v$.
In cyclotron, centripetal force $=$ magnetic force
i.e., $ \frac{m v^{2}}{r}=B q v$
$\therefore \frac{v}{r}=\frac{B q}{m}$
where $\frac{v}{r}=\omega$, therefore
$\omega=\frac{B q}{m}$
and frequency $f=\frac{\omega}{2 \pi}=\frac{B q}{2 m \pi}$
Putting the numerical values from the question, we have
$B=1 T , m=9.1 \times 10^{-31} kg , q=1.6 \times 10^{-19} C $
$\therefore f=\frac{1.6 \times 10^{-19} \times 1}{2 \times 3.14 \times 9.1 \times 10^{-31}}=28 \times 10^{9} Hz $
$\therefore f=28 \,GHz$