= Solution
Normalize the <scale factor> to $a(t_0)=1$ and set $a_d=10^{-3}$. In the same matter-only model $a=(t/t_0)^{2/3}$, so
$$
\boxed{t_d=t_0a_d^{3/2}\simeq2.94\times10^5\ \mathrm{yr},\qquad
r_p(t_d)=3ct_d\simeq0.271\ \mathrm{Mpc}.}
$$
Its present proper size after expansion with the <scale factor> is
$$
\boxed{r_{p,\mathrm{expanded}}=\frac{r_p(t_d)}{a_d}=\frac{2c}{H_0}\sqrt{a_d}\simeq271\ \mathrm{Mpc}.}
$$
The present <comoving distance> to decoupling is
$$
\chi_d=\int_{t_d}^{t_0}\frac{c\,dt}{a(t)}=\frac{2c}{H_0}(1-\sqrt{a_d})\simeq8295\ \mathrm{Mpc}.
$$
The <angular diameter distance> is $D_A=a_d\chi_d$, so the angular size of a transverse length equal to one <particle horizon> radius is
$$
\boxed{\theta\simeq\frac{r_p(t_d)}{D_A}=\frac{\sqrt{a_d}}{1-\sqrt{a_d}}=\frac{1}{\sqrt{1000}-1}\simeq0.0327\ \mathrm{rad}\simeq1.87^\circ.}
$$
If “the horizon” denotes the diameter of the whole causally connected ball, its angular diameter is twice this, approximately $3.74^\circ$. The degree-scale conclusion is the same. Neglecting early <radiation> is an explicit approximation inherited from the question.
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