Answer
$5.9\times 10^{12}$ $m/s^{2}$
Work Step by Step
The magnetic fore acting on a particle having charge $q$ and moving with a velocity $(\vec v)$ in a uniform magnetic field $(\vec B)$ is given by
$ \vec F=q(\vec v\times \vec B)$ ............. $(1)$
Here,
$\vec v \perp \vec B$
and $ |\vec B|=1.6\times10^{-5}$ $T$, $|\vec v|=2.1\times10^{6}$ $m/s $, $q=1.6\times10^{19}C$
Putting the above values in equation $(1)$, we get,
$|\vec F|=qvB$
$|\vec F|=1.6\times10^{-19}\times2.1\times10^{6}\times1.6\times10^{-5}$
Now, $|\vec F|=m_{e}a$, where $m_{e}$ is the mass of electron and $a$ is the magnitude of the acceleration caused by the magnetic force.
Thus, $a=\frac{|\vec F|}{m_{e}}$
or, $a=\frac{1.6\times10^{-19}\times2.1\times10^{6}\times1.6\times10^{-5}}{9.11\times10^{-31}}$ $m/s^{2}$
or, $a=5.9\times 10^{12}$ $m/s^{2}$
Therefore, the magnitude of the acceleration of the electron caused by the magnetic force is $5.9\times 10^{12}$ $m/s^{2}$.