= Solution
In the prescribed matter-dominated approximation, neglect curvature, <radiation>, and the <cosmological constant> throughout the calculation. The <Friedmann equation> and $\rho_m\propto R^{-3}$ give $\dot R\propto R^{-1/2}$, hence $R\propto t^{2/3}$ and $H=2/(3t)$. For $H_0=70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}=2.269\times10^{-18}\,\mathrm{s^{-1}}$,
$$
\boxed{t_0=\frac{2}{3H_0}\simeq9.31\times10^9\ \mathrm{yr}.}
$$
The <particle horizon> is a proper distance on a constant-time slice, not simply $ct$:
$$
r_p(t)=ct^{2/3}\int_0^t(t')^{-2/3}\,dt'=3ct.
$$
Therefore
$$
\boxed{r_p(t_0)=\frac{2c}{H_0}\simeq8.57\times10^3\ \mathrm{Mpc}\simeq27.9\times10^9\ \mathrm{light\ years}.}
$$
These are estimates within a spatially flat, matter-only model, rather than the measured age and <particle horizon> of the actual late-time universe.
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