= Solution
Let $t_*$ be the observation time and normalize $a(t_*)=1$ temporarily. In pure <radiation domination>, $a(t)=(t/t_*)^{1/2}$ and $H_*=1/(2t_*)$. With $p(t)=T_r(t)=T_*/a(t)$, the proper speed is $v=[1+(ma/T_*)^2]^{-1/2}$. Put $u=\sqrt{t/t_*}=a(t)$ in the preceding integral. Since $dt=2t_*u\,du$,
$$
d=2t_*\int_0^1\frac{du}{\sqrt{1+(mu/T_*)^2}}=\boxed{H_*^{-1}\frac{T_*}{m}\operatorname{arsinh}\!\left(\frac m{T_*}\right).}
$$
The small-$m/T_*$ limit is $H_*^{-1}$, the radiation-era <particle horizon>. For $m/T_*\gg1$, the ratio is approximately $(T_*/m)\log(2m/T_*)$, retaining the distance accumulated while the particle was faster.
There is a normalization qualification: in arbitrary fixed <comoving coordinates> the distance is $\chi=H_*^{-1}\operatorname{arsinh}(y)/(a_*y)$, $y=m/T_*$. The expression just derived is also the physical distance $L_*=a_*\chi$ at observation. Thus it equals a comoving distance only when $a_*=1$ is used, as in the preceding part. To express it in today's convention $a_0=1$, multiply $L_*$ by $1+z_*$. This is the <accumulated free-streaming distance in a radiation-dominated universe>.
Write $\omega_M=\Omega_Mh^2$, where $h$ here is the dimensionless present <Hubble constant>. The supplied equality <redshift> gives $1+z_{\rm eq}\simeq2.5\times10^4\omega_M$, and $T_{\rm eq}=T_0(1+z_{\rm eq})$. For $m=30\,\mathrm{eV}$ the requested dimensionless ratio is
$$
\boxed{\frac{L_{\rm eq}}{H_{\rm eq}^{-1}}=\frac{\operatorname{arsinh}(y_{\rm eq})}{y_{\rm eq}},\qquad y_{\rm eq}=\frac{30\,\mathrm{eV}}{T_0(1+z_{\rm eq})}.}
$$
The Hubble rate in this formula is the radiation-era extrapolation used in deriving it. A unique numerical ratio cannot be chosen without $\omega_M$.
\b[The printed <CMB> <temperature> contains a factor-of-ten error.] Its $2.35\,\mathrm{meV}$ means $2.35\times10^{-3}\,\mathrm{eV}$, whereas the <cosmic microwave background temperature in energy units> is about $2.35\times10^{-4}\,\mathrm{eV}=0.235\,\mathrm{meV}$, corresponding to $2.725\,\mathrm K$. The <temperature> is supported by https://asd.gsfc.nasa.gov/archive/arcade/cmb_temperature.html[NASA's CMB measurements]. Using the printed value gives $y_{\rm eq}\simeq0.5106/\omega_M$; using the corrected value gives $y_{\rm eq}\simeq5.106/\omega_M$. For the illustrative choice $\Omega_M=1,h=0.5$, so $\omega_M=0.25$, these give respectively $T_{\rm eq}\simeq14.7$ and $1.47\,\mathrm{eV}$, and distance-to-Hubble-radius ratios about $0.716$ and $0.182$. This benchmark is an assumption for the estimate, not an extra parameter supplied by the paper.
For the present-day comoving length, use $H_r(z_{\rm eq})=H_0\sqrt{\Omega_M}(1+z_{\rm eq})^{3/2}$, the radiation contribution extrapolated to equality. Restoring $c$ for the conversion and using $c/H_0\simeq2998h^{-1}\,\mathrm{Mpc}$ gives
$$
\chi_0\simeq\frac{c}{H_0\sqrt{\Omega_M}\sqrt{1+z_{\rm eq}}}\frac{\operatorname{arsinh}(y_{\rm eq})}{y_{\rm eq}}\simeq\boxed{\frac{18.96}{\omega_M}\frac{\operatorname{arsinh}(y_{\rm eq})}{y_{\rm eq}}\,\mathrm{Mpc}.}
$$
For the same benchmark, this is about $54\,\mathrm{Mpc}$ using the printed <temperature> and $14\,\mathrm{Mpc}$ using the corrected <temperature>, in the prescribed representative-momentum radiation-era model.
At equality matter is no longer negligible, so these are transition-era estimates. The actual total $H_{\rm eq}$ is $\sqrt2H_r(z_{\rm eq})$; inserting it in the simple normalization would lower the length by $\sqrt2$, but would not constitute an exact <integration> through equality. An exact matter-plus-radiation treatment instead replaces the radiation integral by
$$
\chi_0=\frac{c}{H_0\sqrt{\Omega_r}}\int_0^{a_{\rm eq}}\frac{da}{\sqrt{1+a/a_{\rm eq}}\sqrt{1+(ma/p_0)^2}},\qquad a_0=1,
$$
with $p_0=T_0$ in the assumed model. This makes the order-one equality correction explicit. Real <neutrino> <physical momenta in an FRW universe> are distributed and their <temperature> need not equal the <photon> <temperature>; the calculation above follows the specified $p=T_r$ idealization.
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