The number of electrons involved in the electro deposition of 63.5 g. of Cu from a solution of ${\mathrm{CuSO}}_4$ is 

Atomic mass of Cu = 63.5 

Moles of Cu = 1 gmol^-1.

$C{u}^{2+}+\hspace{0.17em}2e^{-}\to Cu$

To deposit  1 mol of Cu , 2 mol of electron required. 

1 mol = 6.023 X 10²³.

to deposit 63.5 g of Cu , 2 X 6.023 X 10²³.electrons are required 

i.e., 12.04 X 10²³. electrons required.