Answer
The number of electron must be transferred from the plate to the rod, so that both objects have the same charge $=1.5625\times10^{13}$ electrons.
Work Step by Step
Given, the charge of the plate, $q_{1}=-3 \mu C$ and the charge of the rod, $q_{2}=+2 \mu C$
Let the amount of charge must be transferred from the plate to the rod, so that both objects have the same charge $=\Delta q$
Now the final charge of of plate would be $=(q_{1}-\Delta)$, which is actually the average of total initial charges.
$(q_{1}-\Delta) =\frac{1}{2}(q_{1}+q_{2})$
$-3 \mu C-\Delta =\frac{1}{2}(-3 \mu C+2 \mu C)$
$-\Delta = -0.5 \mu C+3 \mu C $
$\Delta = -2.5 \mu C $
Charge of an electron $=1.60\times10^{-19} C$
Therefore the number of electron must be transferred from the plate to the rod, so that both objects have the same charge
$=\frac{-2.5\times10^{-6}}{-1.60\times10^{-19}}$
$=1.5625\times10^{13}$
