An electron with energy 0 . 1keV moves at right angle to the...

The frequency of revolution of the electron is $f = \dfrac{eB}{2\pim}$, here, $e$ is charge of electron, $B$ is magnetic field and $m$ is mass of electron.

So, $f = \dfrac{1 . 6 \times {10}^{- 19} \times {10}^{- 4}}{2\pi \times 9 \times {10}^{- 31}} = 2 . 8 \times {10}^{6}Hz$