= Solution
Use time dependence $e^{i\omega t}$ and write the outgoing two-dimensional <Helmholtz equation> kernel as $G_2\sim C e^{-ik_0r}/\sqrt{k_0r}$, where $C$ is independent of the <Mach number>. The harmonic <Lighthill acoustic analogy> gives
$$
\widehat\rho'=\frac1{c_0^2}\partial_i\partial_j\int G_2(\mathbf x-\mathbf y)\widehat T_{ij}(\mathbf y)\,d^2y.
$$
For a compact <acoustic quadrupole>, the source integral is $O(\rho_0U^2\ell^2)$. In the radiation region each derivative of the outgoing exponential supplies a factor of $k_0$, so
$$
\frac{|\widehat\rho'|}{\rho_0}=O\!\left[m^2(k_0\ell)^2(k_0r)^{-1/2}\right].
$$
Keeping the same advective source-frequency scaling as in (i), $\omega=O(U/\ell)$ and hence $k_0\ell=O(m)$. Therefore the <two-dimensional compact quadrupole scaling> is
$$
\boxed{\frac{|\widehat\rho'|}{\rho_0}=O\!\left(m^{7/2}\sqrt{\frac\ell r}\right).}
$$
The changed exponent comes from cylindrical spreading $(k_0r)^{-1/2}$. Compactness $k_0\ell\ll1$ and the far-field condition $k_0r\gg1$ are separate assumptions. At fixed imposed $\omega$ and fixed $\ell$, with only $U$ varied, the stress instead gives an $O(m^2)$ dependence; monochromaticity alone does not supply the additional source-frequency assumption.
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