Answer
${\bf 1.057} \;\rm mA$, ${\bf 12.5}\;\rm T $
Work Step by Step
Assuming that the electron rotates clockwise, so the direction of the current is counterclockwise since we know that the direction of the current is the opposite direction of the direction of the negative charges.
This means that the net magnetic field at the center of the Bohr atom is upward (according to the right-hand rule).
The magnetic field strength of a current loop (as the motion of the electron around the center of the atom) is given by
$$B_{\rm center}=\dfrac{\mu_0I}{2 R}\tag 1$$
Recalling that $I=dq/dt$ where $dq$ here is the charge of the electron itself. So,
$$I_{\rm avg}=\dfrac{dq}{dt}=\dfrac{e}{T}$$
where we know that $v=2\pi R/T$, so $T=2\pi R/v$, and hence,
$$I_{\rm avg }=\dfrac{ev}{2\pi R}\tag 2$$
Plug the known;
$$I_{\rm avg }=\dfrac{(1.6\times 10^{-19})(2.2\times 10^6)}{2\pi (5.3\times 10^{-11})} $$
$$I_{\rm avg }=\color{red}{\bf 1.057} \;\rm mA$$
Plug $I$ from (2) into (1);
$$B_{\rm center}=\dfrac{\mu_0 }{2 R} \cdot\dfrac{ev}{2\pi R}=\dfrac{\mu_0 e v}{4\pi R^2}$$
Plug the known;
$$B_{\rm center} =\dfrac{(4\pi \times 10^{-7})(1.6\times 10^{-19})(2.2\times 10^6)}{4\pi (5.3\times 10^{-11})^2}$$
$$B_{\rm center}=\color{red}{\bf 12.5}\;\rm T\tag{Upward}$$
