= Solution
Using the replacement prescribed in the question, $k_{\rm B}T\to E_F$, gives
$$
\lambda_D^{\rm prescribed}=\sqrt{\frac{\epsilon_0E_F}{n_ee^2}}\simeq6.10\times10^{-11}\,\mathrm m.
$$
The number of electrons in a <Debye sphere> is consequently
$$
\boxed{N_D=\frac{4\pi}{3}n_e\lambda_D^3\simeq0.151.}
$$
A useful check eliminates $n_e$: with $r=\hbar\omega_{\rm pe}/E_F$ and $k_F=(3\pi^2n_e)^{1/3}$, $k_F\lambda_D=\sqrt2/r$ and $N_D=8\sqrt2/(9\pi r^3)$.
Having substantially fewer than one electron in this sphere shows that the classical many-particle Debye-screening picture is not applicable. It does not exclude collective plasmons in a degenerate electron gas. A proper zero-temperature <Thomas-Fermi screening> calculation uses $\partial n/\partial\mu=3n/(2E_F)$ and gives $\lambda_{\rm TF}^2=2\epsilon_0E_F/(3n_ee^2)$, rather than literally replacing $T$ by $T_F$. The boxed count follows the source's stated prescription and exposes its physical limitation.
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