= Solution
Put $r=|\mathbf x|$, $\mathbf n=\mathbf x/r$, and take a <compactly supported> <Lighthill stress tensor>. The isotropic term in that <tensor> is $(p-c_0^2\rho)\delta_{ij}$: the <Kronecker delta> is implicit in the printed shorthand. Convolving the <Lighthill acoustic analogy> with the <retarded acoustic Green function> and integrating twice by parts in the source coordinates gives
$$
\rho'(\mathbf x,t)=\partial_i\partial_j\int\frac{T_{ij}(\mathbf y,t-|\mathbf x-\mathbf y|/c_0)}{4\pi c_0^2|\mathbf x-\mathbf y|}\,d^3y.
$$
The boundary terms vanish because of <compact support>. For source size $\ell$, the <acoustic compact-source approximation> requires $\omega\ell/c_0\ll1$. At $r\gg\ell$, we may therefore replace the denominator by $r$ and the retarded argument by $t-r/c_0$ in the leading source integral. Define $S_{ij}(t)=\int T_{ij}(\mathbf y,t)d^3y$. In the radiation region $\omega r/c_0\gg1$, the leading two spatial derivatives act on the <retarded time> rather than on $1/r$ or $\mathbf n$:
$$
\partial_i\partial_j\frac{S_{ij}(t-r/c_0)}r\sim\frac{n_in_j}{c_0^2r}\ddot S_{ij}(t-r/c_0).
$$
Thus the leading <acoustic quadrupole> density is
$$
\boxed{\rho'(\mathbf x,t)\sim\frac{x_ix_j}{4\pi c_0^4r^3}\ddot S_{ij}(t-r/c_0).}
$$
For the <compact acoustic quadrupole Mach-number scaling>, let $U$ be the source fluctuation speed, $m=U/c_0\ll1$, and use the advective source time $\ell/U$. With $T_{ij}=O(\rho_0U^2)$, one has $S_{ij}=O(\rho_0U^2\ell^3)$ and $\ddot S_{ij}=O(\rho_0U^4\ell)$. Consequently
$$
\boxed{\frac{\rho'}{\rho_0}=O\!\left(m^4\frac\ell r\right).}
$$
The fourth power refers to the radiation coefficient with geometric spreading separated. It assumes that the source <frequency> scales as $U/\ell$; an independently imposed frequency or an independently strong thermal source would change that <Mach number> estimate.
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