= Solution
Let $r_\nu=T_{\nu,0}/T_0$, and retain the assumption that these <neutrinos> remain massless. Constancy of the <neutrino degeneracy parameter> gives $|\mu_{\nu,0}|=|\xi_\nu|r_\nu T_0$. The condition on the <critical density> therefore yields the <critical-density bound on massless neutrino degeneracy>
$$
\frac{|\xi_\nu|^4r_\nu^4T_0^4}{8\pi^2}\lesssim6\times10^3T_0^4,
\qquad
\boxed{|\xi_\nu|\lesssim(48\times10^3\pi^2)^{1/4}r_\nu^{-1}
=26.24\,\frac{T_0}{T_{\nu,0}}}.
$$
For the standard instantaneous-<neutrino decoupling> approximation before <electron-positron annihilation in cosmology>, <cosmological entropy conservation> heats <photons> relative to the decoupled <neutrinos>. The electromagnetic entropy degrees of freedom change from $2+(7/8)4=11/2$ to $2$, giving
$$
T_{\nu,0}=\left(\frac4{11}\right)^{1/3}T_0,
\qquad
\boxed{\left|\frac{\mu_\nu}{T_\nu}\right|\lesssim36.8\ \text{for one flavor}}.
$$
This numerical value requires that thermal-history assumption; the bound in terms of $r_\nu$ is the general answer if large degeneracy changes decoupling or subsequent entropy transfer. Setting $T_{\nu,0}=T_0$ would instead give $26.2$. For $N$ equally degenerate flavors with the same temperature, the leading bound becomes $36.8N^{-1/4}$ under the standard temperature assumption. The neglected thermal corrections make a small change near this loose energy-density limit; this is not a detailed <Big Bang nucleosynthesis> bound.
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