= Solution
In the <canonical ensemble>, $Z(\beta)=\sum_i e^{-\beta E_i}$ normalizes the probabilities $p_i=e^{-\beta E_i}/Z$. For temperature-independent levels,
$$
\boxed{\overline E=-\frac{1}{Z}\frac{dZ}{d\beta}
=-\frac{d\ln Z}{d\beta},\qquad
C=\frac{d\overline E}{dT}
=-\frac{1}{k_BT^2}\frac{d\overline E}{d\beta}.}
$$
The <derivative> holds external parameters, including <magnetic field>, fixed.
For one <independent symmetric three-level system> put $\epsilon=\mu B$ and $x=\beta\epsilon$. Then $Z=1+2\cosh x$ and
$$
\boxed{\overline E=-\frac{2\epsilon\sinh x}{1+2\cosh x}.}
$$
At nonzero fixed field, the ground state dominates as $T\to0$, so $\overline E\to-\mu|B|$. At high <temperature> all three levels become nearly equally populated and their mean tends to zero from below. At fixed positive <temperature> the <energy> is even in $B$, is quadratic near zero field, and approaches $-\mu|B|$ at large $|B|$ as the lowest level becomes occupied. The <heat capacity> vanishes both when excitations are frozen out and when populations are almost equal, with a positive <Schottky anomaly> between those limits.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2014/paper-2-b2-three-level-curves.png]
{title=Three-level mean energy versus temperature and field, and its Schottky heat-capacity peak}
At $|x|\ll1$, expanding the hyperbolic functions gives
$$
\boxed{\overline E\simeq-\frac{2\mu^2B^2}{3k_BT},
\qquad C\simeq\frac{2\mu^2B^2}{3k_BT^2}.}
$$
Interpreting the stated <heat capacity> on the same one-atom basis as the preceding calculation, its tail coefficient gives
$$
\boxed{\mu=\frac1B\sqrt{\frac{3ak_B}{2}}
\simeq2.89\times10^{-5}\,\mathrm{eV\,T^{-1}}
=4.64\times10^{-24}\,\mathrm{J\,T^{-1}}\simeq0.500\,\mu_B.}
$$
Here $k_B=8.61733\times10^{-5}\,\mathrm{eV\,K^{-1}}$ is used consistently with $a$ in electronvolt kelvin. For a sample of $N$ independent atoms, the total coefficient would be $N$ times the one-atom coefficient; it must first be divided by $N$.
Finally, two derivatives of $Z$ supply the second <energy> moment:
$$
\overline{E^2}=\frac{Z''}{Z},\qquad
\frac{d\overline E}{d\beta}
=-\left(\frac{Z''}{Z}-\frac{Z'^2}{Z^2}\right)
=-\operatorname{Var}(E).
$$
Combining this with the heat-capacity <derivative> yields the <canonical energy fluctuation> relation
$$
\boxed{C=\frac{\operatorname{Var}(E)}{k_BT^2},
\qquad\Delta E=\sqrt{k_BT^2C}.}
$$
It also explains why this canonical <heat capacity> is nonnegative.<Statistical Physics by David Tong>[§§1.3.1–1.3.2, printed pages 18–20, equations (1.25)–(1.27)]{ref}
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