Each atom of an iron bar (5cm x 1cm x 1crn) has a magnetic moment $1.8 \mathrm{x} {10}^{-23} {\mathrm{Am}}^2$. Knowing that the density of iron is $7.78 \mathrm{x}{10}^3 \mathrm{kg}{m}^{-3}$, atomic weight is 56 and Avogadro's number is $6.02 \mathrm{x} {10}^{23}$, the magnetic moment of bar in the state of magnetic saturation will be

The number of atoms per unit volume in a specimen, 

$\mathrm{n}=\frac{{\mathrm{\rho N}}_{\mathrm{A}}}{\mathrm{A}}$

For iron, $\mathrm{\rho }=7.78 \mathrm{x} {10}^3 {\mathrm{kgm}}^{-3},$

${\mathrm{N}}_{\mathrm{A}}=6.02 \mathrm{x} {10}^{26} /\mathrm{kgmol}, \mathrm{A}=56$

$\Rightarrow \mathrm{n}=\frac{7.78 \mathrm{x} {10}^3 \mathrm{x} 6.02 \mathrm{x} {10}^{26}}{56}=8.38 \mathrm{x} {10}^{28} {\mathrm{m}}^{-3}$

Total number of atoms in the bar is

${\mathrm{N}}_0=\mathrm{nV}=8.38 \mathrm{x} {10}^{28} \mathrm{x} \left(5 \mathrm{x} {10}^{-2} \mathrm{x} 1 \mathrm{x} {10}^{-2} \mathrm{x} 1 \mathrm{x} {10}^{-2}\right)$

${\mathrm{N}}_0=4.19 \mathrm{x} {10}^{23}$

The saturated magnetic moment of bar

$=4.19 \mathrm{x} {10}^{23} \mathrm{x} 1.8 \mathrm{x} {10}^{-23}=7.54 {\mathrm{Am}}^2$