= Solution
For interface area $\mathcal A$, each resolved spin branch has one state per flux increment $h/e$, so its <degeneracy of a Landau level> is
$$
N_\phi=\frac{B\mathcal A}{h/e},\qquad \frac{N_\phi}{\mathcal A}=\frac{eB}{h}.
$$
If $\nu$ such branches are occupied, including fractional occupation of the highest, the <filling factor> gives
$$
\boxed{n_s=\frac{N}{\mathcal A}=\nu\frac{eB}{h}.}
$$
The filling counts spin and any other internal branches separately; it is an integer when a whole number of resolved levels are full.
From the visible PDF axes, approximately ten unsplit periods lie between $1/B=1.0$ and $2.5\,\mathrm{T^{-1}}$, giving $\Delta(1/B)\simeq0.15\,\mathrm{T^{-1}}$. Consecutive unsplit orbital peaks differ by two occupied spin branches, so <carrier density from spin-degenerate quantum oscillations> gives
$$
\boxed{n_s=\frac{2e}{h\Delta(1/B)}
\simeq3.2\times10^{15}\,\mathrm{m^{-2}}
=3.2\times10^{11}\,\mathrm{cm^{-2}}.}
$$
The precision is limited by reading the plotted trace, roughly a few percent. The factor two assumes only spin remains unresolved in that part of the graph; counting individual spin-split peaks would instead use a one-branch filling increment.
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