= Solution
Write $\delta=\delta_{C,\mathbf k}$ and $\mathcal H=a'/a$, the <conformal Hubble parameter>. The flat <Friedmann equation> gives
$$
\mathcal H^2=\frac{8\pi G}{3}a^2(\rho_C+\rho_R).
$$
Outside the horizon, the stated <adiabatic cosmological perturbation> relation is $\delta_R=(4/3)\delta$. It expresses equality of $\delta_i/(1+w_i)$ for dust and radiation in the density convention of the supplied equations. During <radiation domination>, $a\propto\tau$, so $\mathcal H=1/\tau$ and $\rho_C$ is negligible. Consequently $8\pi Ga^2\rho_R=3/\tau^2$, and the <linear cosmological density perturbation> equation becomes
$$
\delta''+\frac1\tau\delta'=\frac4{\tau^2}\delta.
$$
The trial solution $\delta\propto\tau^q$ gives $q(q-1)+q=4$, hence $q=2,-2$.
During <matter domination>, $a\propto\tau^2$, so $\mathcal H=2/\tau$. Now the radiation source is negligible and $4\pi Ga^2\rho_C=(3/2)\mathcal H^2=6/\tau^2$. The equation becomes
$$
\delta''+\frac2\tau\delta'=\frac6{\tau^2}\delta,
$$
whose exponents satisfy $q(q-1)+2q=6$, or $(q-2)(q+3)=0$. Thus the growing solutions in both eras are
$$
\boxed{\delta_{C,\mathbf k}\propto\tau^2.}
$$
The other modes are $\tau^{-2}$ in the radiation limit and $\tau^{-3}$ in the matter limit. These are statements in the density and gauge convention of the supplied evolution equation; a density contrast outside the horizon is not a gauge-independent observable by itself.
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