= Solution
Take dimensionless <spin> operators and $H=DS_z^2+g\mu_BBS_z$, with $g>0$. The four <wavefunctions> are $|m\rangle$, $m=-3/2,-1/2,1/2,3/2$, with
$$
\boxed{E_m=Dm^2+g\mu_BBm}.
$$
At zero field they form doublets at $D/4$ and $9D/4$. A transverse microwave <magnetic field> drives $\Delta m=\pm1$ transitions. When $h\nu>2D$, their positive resonance fields in ascending order are
$$
B_1=\frac{h\nu-2D}{g\mu_B},\quad B_2=\frac{h\nu}{g\mu_B},\quad B_3=\frac{h\nu+2D}{g\mu_B}.
$$
They connect $(1/2,3/2)$, $(-1/2,1/2)$, and $(-3/2,-1/2)$, respectively. Thus
$$
\boxed{g=2.382,\quad D=0.425h\nu=2.82\times10^{-24}\ \mathrm J=0.0176\ \mathrm{meV},\quad B_3=0.555\ \mathrm T}.
$$
Here $D/h=4.25\ \mathrm{GHz}$ and the zero-field doublet gap is $2D$.
At lower frequency it is essential to use absolute energy gaps, not discard a transition after its levels cross. For $0<h\nu'<2D$, the distinct candidate positive fields are
$$
\frac{h\nu'}{g\mu_B},\quad \frac{2D-h\nu'}{g\mu_B},\quad \frac{2D+h\nu'}{g\mu_B}.
$$
The first two coincide when
$$
\boxed{\nu'=D/h=4.25\ \mathrm{GHz}}.
$$
This <axial spin-three-halves resonance coincidence> leaves two distinct resonance positions, at $0.1275$ and $0.3825\ \mathrm T$, with two transitions superposed at the lower position. At $\nu'=2D/h$ there are also only two \i[strictly positive-field] lines, but an additional zero-field resonance remains; this is a different counting convention. Observable intensities also require thermal occupation of the relevant lower levels.
\Image[/past-exam-of-the-condensed-matter-physics-major-option-course-of-the-university-of-oxford/2016-spin-three-halves-resonances.png]
{title=Axial spin-three-halves levels and the three high-frequency resonance fields}
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