Answer
$4.27\times 10^{-5}\;rad/s$
Work Step by Step
The iron sphere initially had no net magnetic moment. Under external magnetic field, the magnetic moment arises due to the spin angular momentum of the aligned loosely bound electrons. Each atom has one loosely bound electron. If $m$ be the mass of the sphere, then the number of loosely bound electrons in the iron is given by
$n=\frac{mN_A}{M}$, where $N_A$ is the Avogadro's number and $M$ is the molar mass of iron.
Thus, total
As the field aligns $12\%$ of the loosely bound electrons, the total spin angular momentum of the loosely bound electrons is
$L_e=\frac{0.12mN_A}{M}(m_s\hbar)$
If $I$ be the moment of inertia, and $\omega$ is the angular speed of the sphere, then the angular momentum of the sphere is
$L_s=I\omega$
In the given condition,
$L_s=L_e$
or, $I\omega=\frac{0.12mN_A}{M}(m_s\hbar)$
or, $\omega=\frac{0.12mN_A}{IM}(m_s\hbar)$
or, $\omega=\frac{0.12mN_A}{(\frac{2}{5}mR^2)M}(m_s\hbar)$
or, $\omega=\frac{0.6N_A}{2R^2M}(m_s\hbar)$
Substituting given values,
$\omega=\frac{0.6\times 6.022\times 10^{23}}{2\times(0.002)^2\times0.0558}(\frac{1}{2}\times\frac{6.63\times10^{-34}}{2\pi})\;rad/s$
$\implies \omega\approx 4.27\times 10^{-5}\;rad/s$
$\therefore$ The angular speed of the sphere is $4.27\times 10^{-5}\;rad/s$.
