= Solution
{c}
Add the electronic <diamagnetism> of each helium atom to the nuclear <magnetic susceptibility>, using the same $n_3$ for both contributions in the <helium-3> sample:
$$
\chi_{3,\mathrm{tot}}\simeq\mu_0n_3\left[\frac{(2.13\mu_N)^2}{k_BT}-\frac{e^2\langle r^2\rangle}{3m_e}\right].
$$
The density cancels when the two terms balance. The sign-change temperature in the independent-atom approximation is therefore
$$
\boxed{T_0=\frac{3m_e(2.13\mu_N)^2}{e^2k_B\langle r^2\rangle}.}
$$
No numerical atomic radius was supplied, so the exact numerical value remains conditional on that electronic length scale. Taking $\langle r^2\rangle\sim a_0^2$, where $a_0$ is the <Bohr radius>, gives $T_0\simeq0.32\,\mathrm K$. A slightly more explicit estimate uses the <helium variational effective charge> $\zeta=27/16$, for which a hydrogenic $1s$ orbital has $\langle r^2\rangle=3a_0^2/\zeta^2\simeq1.05a_0^2$, giving
$$
\boxed{T_0\approx0.3\,\mathrm K\quad\text{as an atomic-model estimate}.}
$$
Above this scale the electronic contribution is negative and dominates; below it the positive nuclear contribution dominates within the independent-moment approximation. The estimate is not a precision prediction of a real interacting helium phase.
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