In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0 . 5   A…

Time taken to revolve in circular orbit is $T=\frac{2\pi }{\mathrm{\omega }}=\frac{2\pi r}{v}=\frac{2\times {\displaystyle \frac{22}{7}}\times 5\times {10}^{-11}}{2.2\times {10}^6}=\frac{10}{7}\times {10}^{-16} \mathrm{s}$

Now, current associated with the electron is $I=\frac{e}{T}=\frac{1.6\times {10}^{-19}}{{\displaystyle \frac{10}{7}}\times {10}^{-16}}=1.12\times {10}^{-3}\mathrm{A}$
$\Rightarrow I=1.12 \mathrm{mA}$
Hence, the current is $112\times {10}^2 \mathrm{mA}$.