= Solution
Count one neutrino species as the usual two thermally populated particle-antiparticle helicity states, with negligible chemical potential. Its entropy weight is $g_{*s,X}=(7/8)2=7/4$. In the ideal high-temperature <Standard Model>, the bosonic count is $16+8+4=28$ from <gluons>, electroweak <gauge bosons> and the <Higgs field>. The fermionic count is $72+12+6=90$ from <quarks>, charged <leptons> and the three ordinary <neutrino> species. Hence the <relativistic entropy content of the Standard Model> is $g_{*s,D}=28+(7/8)90=106.75$ for the remaining bath. Before the new species leaves, the total count is $108.5$; its separately cooling states are excluded from the subsequent bath count.
Near ordinary <neutrino decoupling>, the still-coupled bath has
$$
g_{*s}=2+\frac78(4+6)=10.75
$$
from photons, electron-positron pairs and the three ordinary neutrinos. After those neutrinos decouple, the electromagnetic bath falls from $2+(7/8)4=11/2$ to two when the pairs annihilate. <Cosmological entropy conservation> in that bath gives $T_\nu/T_\gamma=(4/11)^{1/3}$.
The new species has cooled as $T_X\propto a^{-1}$ since its earlier decoupling, whereas annihilation of Standard Model species heated the remaining bath. The <temperature ratio of an early-decoupled neutrino> is thus
$$
\frac{T_X}{T_\nu}=\left(\frac{10.75}{106.75}\right)^{1/3},\qquad
\boxed{\frac{T_X}{T_\gamma}=\left(\frac{43}{11\times106.75}\right)^{1/3}\simeq0.332.}
$$
These are ideal-gas estimates with the high-temperature count implicit in the question; precise electroweak thresholds change them slightly.
Entropy degrees of freedom weight temperature ratios with their cube, while energy degrees of freedom use their fourth power. The remaining Standard Model entropy today, in photon-temperature units, is $g_{*s,0}^{\mathrm{SM}}=2+(7/8)6(4/11)=43/11$. The new contribution is $(7/4)(T_X/T_\gamma)^3$. Therefore the <fractional entropy of a relativistically decoupled species> is
$$
\boxed{\frac{s_X}{s_{\mathrm{total}}}
=\frac{(7/4)(T_X/T_\gamma)^3}{43/11+(7/4)(T_X/T_\gamma)^3}
=\frac{7/4}{106.75+7/4}\simeq0.0161.}
$$
About \b[1.6% of the total entropy] is in the new species. Separate comoving entropy conservation also derives this fraction directly at decoupling. A model populating four helicity states instead would require the corresponding doubled species count.
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