Answer
See the detailed answer below.
Work Step by Step
We know that the angular momentum of a particle is given by
$$L=mrv\tag 1$$
and since the electron orbits in a 5 mT field. So the electron undergoes a cyclotron motion in which the radius is given by
$$r=\dfrac{mv}{qB}$$
Multiplying both sides by $4r$,
$$4r^2=\dfrac{4mvr}{qB}=\dfrac{4L}{qB}$$
where the diameter of the orbits is $D=2r$, so $4r^2=D^2$, and hence
$$D =2\left[ \sqrt{\dfrac{ L}{qB}}\;\right]$$
Plug the known;
$$D =2\left[ \sqrt{\dfrac{ (8\times 10^{-26})}{(1.6\times 10^{-19})(5\times 10^{-3})}}\;\right]$$
$$D=\color{red}{\bf 2.0}\;\rm cm$$
