= Solution
Normalize $a(t_0)=1$ and put $k_0=2\pi/(ct_0)$. The <cosmological horizon crossing> condition is
$$
\frac{t_H}{t_0}=\frac{k_0}{k}a(t_H).
$$
For $t_H>t_{\rm eq}$, matter domination gives $a(t_H)=(t_H/t_0)^{2/3}$, and therefore
$$
\boxed{\frac{t_H}{t_0}\simeq\left(\frac{k_0}{k}\right)^3.}
$$
For $t_H<t_{\rm eq}$, radiation domination and continuity at equality give
$$
a(t)=a_{\rm eq}\left(\frac{t}{t_{\rm eq}}\right)^{1/2},
\qquad
a_{\rm eq}=\left(\frac{t_{\rm eq}}{t_0}\right)^{2/3}
=\frac1{1+z_{\rm eq}}.
$$
Substitution in the crossing condition yields
$$
\boxed{
\frac{t_H}{t_0}\simeq
\frac1{\sqrt{1+z_{\rm eq}}}
\left(\frac{k_0}{k}\right)^2.}
$$
These are the two branches of the <horizon-crossing time across matter-radiation equality>.
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