= Solution
For $B=0.05$ T, use the two <Spin-only magnetic moment>[spin-only] moment states. Their probabilities and normalized <magnetization> are
$$
p_+=\frac{e^x}{2\cosh x}=\frac{1+\tanh x}{2},\quad
p_-=1-p_+,\quad \frac{M}{M_s}=p_+-p_-=\tanh x,
\quad x=\frac{\mu_B B}{k_BT}.
$$
At 300 K, $x=1.11952\times10^{-4}$; at 2 K, $x=0.0167928$. Therefore
$$
\boxed{\begin{array}{c|cc}
T&\text{aligned population }p_+&\text{net polarization }M/M_s\\\hline
300\ \mathrm K&0.50005598&1.11952\times10^{-4}\\
2\ \mathrm K&0.50839563&0.0167913
\end{array}}
$$
The phrase “fraction aligned” is ambiguous if it is used to mean net alignment. Half the moments are already in the parallel state at zero field, but the net <magnetization> is then zero because the antiparallel half cancels them. The table reports both the literal parallel-state population and the imbalance that produces <magnetization>. At this weak field even the 2 K sample is far from saturation, so the <Curie's law>[Curie approximation] is accurate within this independent-spin model.
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