= Solution
Using the magnitude of the classical slope,
$$
n_s=\frac{1}{e(4000\,\mathrm{m^2/C})}
=1.56\times10^{15}\,\mathrm{m^{-2}}.
$$
The <Landau filling factor> at $1\,\mathrm T$ is
$$
\nu_{\mathrm{fill}}=\frac{n_sh}{eB}
=6.47.
$$
Under the convention that the “last Landau level” means the last resolved spin branch, its degeneracy per area is $eB/h$. Setting $\nu_{\mathrm{fill}}=1$ gives
$$
B_1=\frac{n_sh}{e}=6.47\,\mathrm T.
$$
Above this threshold all carriers fit into that lowest resolved branch, which is partially filled when $\nu_{\mathrm{fill}}<1$.
There is a spin convention to state explicitly: if “Landau level” instead means one orbital level with unresolved twofold spin degeneracy, its capacity is $2eB/h$, and all carriers first fit into orbital index $n=0$ at \b[$B=3.24\,\mathrm T$], where $\nu_{\mathrm{fill}}=2$. The slope alone determines $n_s$ and $\nu_{\mathrm{fill}}$, but does not specify the spin splitting.
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