Explanation

(c) When A is area of cross-section of wire, and n be number of free electrons per unit volume, then relation between electric current (i) and drift velocity ($v_{d}$) is
$i = ne\ Av_{d}$
Number of atoms in 63 g of copper is equal to Avogadro’s number ie, $6 \times 10^{23}$
Volume of 63 g copper
$= \frac{63}{density} = \frac{63}{9} = 7cm^{3}$
$n = \frac{6.02 \times 10^{23}}{7}\ per\ cm^{3}$
$= \left( \frac{6.02 \times 10^{29}}{7} \right)per\ m^{3}$
$Area\ A = \pi r^{2}$
$= \pi\left( 0.5 \times 10^{- 3} \right)^{2}m^{2}$
Hence, drift velocity
$v_{d} = \frac{1.1}{\left( \frac{6.02 \times 10^{29}}{7} \right) \times 1.6 \times 10^{- 19} \times \pi \times \left( 0.5 \times 10^{- 3} \right)^{2}}$
$v_{d} = 0.1 \times 10^{- 3}ms^{- 1}$
$v_{d} = 0.1\ mms^{- 1}$