= Solution
Let $n=N/\mathcal V$ and $c_e=C_e/\mathcal V$ be the <carrier density> and electronic <heat capacity> per unit volume. In isotropic <kinetic theory>, the <angular average> $\langle v_x^2\rangle=v_F^2/3$ gives the electronic <thermal conductivity>
$$
K_e=\frac13c_ev_F\ell_Q=\frac13c_ev_F^2\tau_Q,
$$
where $\ell_Q=v_F\tau_Q$ is the <mean free path> associated with the <heat-current relaxation time>. With $v_F^2=2E_F/m^*$,
$$
K_e=\frac{2A}{3}\frac{nk_B^2T\tau_Q}{m^*}.
$$
The <Drude model> gives the <electrical conductivity> $\sigma=ne^2\tau_\sigma/m^*$, where $\tau_\sigma$ is the <momentum relaxation time>. Therefore
$$
\frac{K_e}{\sigma T}=\frac{2A}{3}\left(\frac{k_B}{e}\right)^2\frac{\tau_Q}{\tau_\sigma}.
$$
When the same approximately elastic collisions relax heat and charge currents, $\tau_Q=\tau_\sigma$. Using $A=\pi^2/2$ gives the <Wiedemann-Franz law>
$$
\boxed{K_e=L_0\sigma T,\qquad L_0=\frac{\pi^2}{3}\left(\frac{k_B}{e}\right)^2
\simeq2.44\times10^{-8}\,\mathrm{W\,\Omega\,K^{-2}}.}
$$
$L_0$ is the <Lorenz number>. Degenerate electrons move at the <Fermi velocity>, not the classical thermal velocity; using classical equipartition would give the wrong coefficient.
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