= Solution
The <isospin> coupling $1\otimes1\to1$, with third component $-1$, has the normalized state
$$
|M^-\rangle=\frac{|\rho^0\pi^-\rangle-|\rho^-\pi^0\rangle}{\sqrt2}.
$$
The two <Clebsch-Gordan coefficients> have equal squared magnitude. Neglecting electromagnetic effects and charged-neutral <mass> differences, the two <rho meson> charge channels have equal rates. Their subsequent strong decays are $\rho^0\to\pi^+\pi^-$ and $\rho^-\to\pi^-\pi^0$, with nearly unit probabilities. Consequently
$$
\boxed{\frac{\Gamma(M^-\to\pi^-\pi^+\pi^-)}{\Gamma(M^-\to\pi^-\pi^0\pi^0)}=1}
$$
in the ideal <isospin> model.
Each final state contains one pair of identical <pions>. Summing the alternative resonant-pair assignments coherently, with the corresponding identical-particle <phase space> factors, does not generate an extra factor of two for just one channel. In the narrow-<rho meson> approximation the equality is already evident from equal $\rho\pi$ rates; the consistent <isospin> amplitudes extend it to the integrated three-body distributions.
The tabulated ratio $9.02/9.30=0.970$ is close to this expectation but is not exactly one. Treating the printed errors as independent gives an uncertainty about $0.012$, so a percent-level comparison requires <isospin breaking>, radiative corrections, realistic resonant dynamics and experimental correlations. The idealized argument predicts the leading pattern rather than an exact identity for the rounded measurements.
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