Solution :
Calculate the drift velocity $V_d= \large\frac{I}{neA}$
Given $n= 85 \times 10^{28}/m^3$
$A= 2 \times 10^{-6}\;m^{2}$
$I= 3A$
$e= 1.6 \times 10^{-7} \;C$
Time taken by electron to drift from one end to continue of the wire,
$t=\large\frac{\text{length of the wire}}{\text{drift velocity}}$
$\quad= \large\frac{l}{V_d}$
Use the relation ,
$I= ne AV_d$
=> $ V_d= \large\frac{I}{n_eA}$
$t=\large\frac{l n_e A}{I}$
substituting the values we get,
$t= \large\frac{ln_eA}{I}$
$\quad= \large\frac{3 \times 8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}{3}$
$\therefore t= 2.72 \times 10^4\;S$
$\qquad=$7 hours and minute
The time taken by an electron to drift from one end to another end is 7 hours and 33 mins