= Solution
Here $H=\dot a/a=2/(3t)$, so the perturbation equation becomes $\ddot\delta+(4/3t)\dot\delta-(2/3t^2)\delta=0$. A trial $t^s$ gives $s(s-1)+(4/3)s-2/3=0$, with roots $2/3,-1$. Thus
$$
\boxed{\delta_k(t)=C_k t^{2/3}+D_k t^{-1}.}
$$
The pure growing mode obeys $\delta_k(t)/\delta_k(t_i)=a(t)/a(t_i)$; for generic data the decaying component eventually becomes negligible, though the growing-mode normalization must be determined from both initial data.
With $a(t_0)=1$, the physical wavelength is $2\pi a(t)/k$. Equating it to the <Hubble radius> $c/H=3ct/2$, and using $k_0=2\pi H_0/c$, gives
$$
\frac{k_0}{k}\left(\frac{t_k}{t_0}\right)^{2/3}
=\frac{t_k}{t_0},\qquad
\boxed{t_k/t_0=(k_0/k)^3.}
$$
Growing-mode evolution from crossing to today multiplies the perturbation amplitude by $(t_0/t_k)^{2/3}=(k/k_0)^2$. Hence the supplied horizon-crossing variance gives
$$
\boxed{P(k)=\frac{A}{k^3}\left(\frac{k}{k_0}\right)^4
=\frac{A}{k_0^4}k.}
$$
This conclusion uses the stated matter-dominated growing-mode approximation over the relevant evolution, rather than an additional radiation-era transfer function.
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