= Solution
Use positive <angular frequencies> and abbreviate $p=\omega_{pe}$, $b=\omega_{ce}$. The left-handed index is $n_L^2=1-p^2/[\omega(\omega+b)]$. It is negative below its positive zero,
$$
\boxed{\omega_L=\frac{\sqrt{b^2+4p^2}-b}{2}.}
$$
For the right-handed mode, $n_R^2=1-p^2/[\omega(\omega-b)]$. Below the <cyclotron resonance> $0<\omega<b$ its denominator is negative and the index square is positive. Just above the resonance it is negative until
$$
\boxed{\omega_R=\frac{b+\sqrt{b^2+4p^2}}2,\qquad
b<\omega<\omega_R\text{ is the stop band}.}
$$
This is the lossless electron-only cold-plasma model. <Ions>, collisions and resonance absorption matter outside its applicable range.
For a whistler with $\omega\ll b$ and $p^2/(\omega b)\gg1$,
$$
k\simeq\frac p{c\sqrt b}\sqrt\omega,\qquad
v_g=\left(\frac{dk}{d\omega}\right)^{-1}
\simeq\frac{2c\sqrt{b\omega}}p.
$$
The travel <time> through a uniform slab is $t-t_s=Lp\,\omega^{-1/2}/(2c\sqrt b)$. In terms of ordinary <frequency>,
$$
\boxed{\frac{d(f^{-1/2})}{dt}=\frac{2c\sqrt{2\pi b}}{Lp}.}
$$
An arbitrary emission <time> changes the intercept, not the slope. The original figure labels a dispersion constant $D=81\pm0.5\,\mathrm{Hz^{1/2}\,s}$: it is the reciprocal of the plotted slope, not the slope itself. Therefore
$$
p=\frac{2c\sqrt{2\pi b}}L D,
\qquad b=\frac{eB}{m_e}.
$$
For $B=80$ nT and $L=6.0\times10^7$ m this gives $b\simeq1.41\times10^4\,\mathrm{s^{-1}}$, $p\simeq2.41\times10^5\,\mathrm{s^{-1}}$, and
$$
\boxed{n_e=\frac{\varepsilon_0m_ep^2}{e^2}\simeq1.8\times10^7\,\mathrm{m^{-3}}
\simeq18\,\mathrm{cm^{-3}}.}
$$
The resulting forbidden right-handed <frequency> width is
$$
\boxed{\Delta f=\frac{\omega_R-b}{2\pi}
=\frac{\sqrt{b^2+4p^2}-b}{4\pi}\simeq3.7\times10^4\,\mathrm{Hz}.}
$$
The lower edge is about $2.24$ kHz and the upper edge about $39.5$ kHz. In the real nonuniform <plasma>, the travel-time coefficient measures a path <integral> proportional to $\int\omega_{pe}/\sqrt{\omega_{ce}}\,ds$. The inferred <density> is consequently an effective uniform-slab estimate, not a unique local <density> at the spacecraft. Uncertainty in the path and <magnetic field> likely exceeds the formal error in $D$.
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