= Solution
Normalize the <Einstein-de Sitter universe> by $a(t_0)=1$, giving $a(t)=(t/t_0)^{2/3}$. A photon emitted inward today at the current <comoving particle horizon> starts at $\chi_0=3ct_0$. By time $t$ it has traversed the comoving distance
$$
\int_{t_0}^{t}\frac{c\,dt'}{a(t')}=3ct_0\left[\left(\frac{t}{t_0}\right)^{1/3}-1\right].
$$
At arrival at Earth this equals $\chi_0$, hence $(t/t_0)^{1/3}=2$ and
$$
\boxed{t_{\mathrm{arrival}}=8t_0,\qquad t_{\mathrm{travel}}=7t_0.}
$$
The emitting point is at the horizon location today, not a photon already arriving today from the initial singularity. The calculation of <photon arrival from an Einstein-de Sitter particle horizon> follows the future inward <null geodesic> from that event.
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