In Bohr's atomic model, the electron is assumed to revolve in a...

Time taken to revolve in circular orbit is $T = \dfrac{2\pi}{\omega} = \dfrac{2\pir}{v} = \dfrac{2 \times \dfrac{22}{7} \times 5 \times {10}^{- 11}}{2 . 2 \times {10}^{6}} = \dfrac{10}{7} \times {10}^{- 16}s$

Now, current associated with the electron is $I = \dfrac{e}{T} = \dfrac{1 . 6 \times {10}^{- 19}}{\dfrac{10}{7} \times {10}^{- 16}} = 1 . 12 \times {10}^{- 3}A$
$⇒ I = 1 . 12mA$
Hence, the current is $112 \times {10}^{2}mA$.