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Use the flat matter-only Einstein-de Sitter approximation for the distance-redshift relation, neglecting radiation corrections at cosmological recombination. Let $z_{\ast }$ be the cosmological recombination redshift. Adiabatic photon cooling gives $1+z_{\ast }=T_{\ast }/T_0$. The Hubble rate is $H(z)=H_0(1+z{)}^{3/2}$, so the comoving radial distance to last scattering isThe angular diameter distance is $D_A={\chi }_{\ast }/(1+z_{\ast })$. A proper Hubble radius $H_{\ast }^{-1}=H_0^{-1}(1+z_{\ast }{)}^{-3/2}$ therefore subtends the small angleThis is the angular Hubble-radius scale in an Einstein-de Sitter universe; the present Hubble constant cancels.

Taking $T_{\ast }=0.3\mathrm{eV}$ and the literal printed $T_0=2.35\mathrm{meV}$ gives $1+z_{\ast }=127.7$ and ${\theta }_H\simeq 0.0486$ radians, or $2.8$ degrees. As noted above, that present CMB energy is too large by ten. With the corrected $T_0=0.235\mathrm{meV}$, $1+z_{\ast }=1277$ andwhich gives the familiar order-one-degree scale within the assumed cosmology. Using $0.28\mathrm{eV}$ to mark a few-per-cent ionization fraction changes this only modestly. These numbers refer to the Hubble-radius scale, not its diameter. In a matter-only universe the particle-horizon radius is $2H^{-1}$ and would give twice the angle; the acoustic sound horizon is a different length again.