= Solution
At $B_z=10\,\mathrm T$, the nominal filling is
$$
\boxed{\nu_{10}=\frac{N_Sh}{10e},\qquad
E_{n,\sigma}=\hbar\frac{eB_z}{m^*}(n+\tfrac12)+\sigma\frac{g^*\mu_BB_z}{2}.}
$$
A labelled density-of-states picture has disorder-broadened peaks at these orbital and <spin> energies, localized tails and extended states near their centres. A <chemical potential> in the localized region corresponds to a Hall plateau and small <longitudinal resistivity>; passing through an extended-state centre corresponds to a transition.
\b[TODO — missing density/plateau assignment:] there is no basis for choosing one of those positions at 10 T without the missing Hall graph. For example $N_S=4.84\times10^{15}\,\mathrm{m^{-2}}$ gives nominal $\nu_{10}\simeq2$, whereas $N_S=6.04\times10^{15}\,\mathrm{m^{-2}}$ gives $\nu_{10}\simeq2.5$. In an ideal resolved-level picture the first can put the <chemical potential> between filled levels, while the second leaves a level partly filled. Both are compatible with the surviving current and width, demonstrating why the requested placement is undetermined.
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