= Solution
{c}
Measure <magnetization> along $z$ at low temperature and use a small field to select a single weak-ferromagnetic domain. After domain alignment, subtract the approximately linear field-induced susceptibility background to obtain the spontaneous weak moment, using the extrapolated intercept of that branch rather than a multidomain zero-field average.
There are four iron ions per orthorhombic cell. If the spontaneous magnetization density is $M_z$ and the cell volume is $V_{\mathrm{cell}}$, the ordered moment and dimensionless spin component per ion are
$$
\boxed{m_z=\frac{M_zV_{\mathrm{cell}}}{4}=g\mu_BS\sin\theta,\qquad\langle S_z\rangle=\frac{M_zV_{\mathrm{cell}}}{4g\mu_B}.}
$$
Here spin components are expressed in units of $\hbar$ as in the source Hamiltonian. Alternatively normalize the magnetic moment to the measured number of iron ions. A field too large to preserve the small-canting structure would not measure the zero-field component directly. For high-spin Fe$^{3+}$ with $S=5/2$ and $g\simeq2$, the angle above corresponds to approximately $0.026\mu_B$ per iron ion.
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