Answer
$3.2\times 10^{15}\;\rm electron$
Work Step by Step
The plate on the ceiling will pull you up when the electric attractive force between your body and the plate is just equal to your weight.
Hence,
$$F_G=F_{electric}\tag 1$$
The plate is losing electrons that are fired toward you, so the plate is building up a positive charge while your body is receiving the electrons from the electron pump.
This means that the final positive charge in the plate is equal to the final negative charge in your body, $q_1=q_2=q$
Hence, from (1)
$$m_{you}g=\dfrac{kq^2}{r^2}$$
Thus, the charge in your body is
$$q=\sqrt{\dfrac{m_{you}g }{k}}\;r\tag 2$$
Now we need to find $q$, then find the number of electrons which is given by dividing the net charge by the charge of one electron.
$$N_e=\dfrac{q}{q_{e^-}}$$
Plug from (2);
$$N_e=\dfrac{\sqrt{\dfrac{m_{you}g }{k}}\;r}{q_{e^-}}$$
Plugging the known;
$$N_e=\dfrac{\sqrt{\dfrac{(60)(9.8) }{8.99\times 10^9}}\times (2)}{(1.602 \times 10^{-19})}$$
Note that the author told us to treat both, you and the plate, as point charges. So the distance between you and the plate is 2 m.
$$N_e=\color{red}{\bf 3.2\times 10^{15}}\;\rm electron$$
