Each atom of an iron bar $(5 \, cm \times 1 \, cm \times 1 \, cm)$ has a magnetic moment of $1.8 \times 10^{-23} \, A \cdot m^2$. Given that the density of iron is $7.78 \times 10^3 \, kg/m^3$,the atomic weight is $56 \, g/mol$,and Avogadro's number is $6.02 \times 10^{23} \, mol^{-1}$,calculate the magnetic moment of the bar in the state of magnetic saturation in $A \cdot m^2$.

• A
  $4.75$
• B
  $5.74$
• C
  $7.54$
• D
  $75.4$