= Solution
Use $c=1$ and let $E$ denote the \b[total] center-of-momentum energy, not one beam's energy. Opposite equal momenta give $E=2E_{\rm beam}=498$ MeV, so the <kaon> is at its stipulated resonance mass $m_K=498$ MeV. Write the unspecified selected exit <branching fractions> as $b_\rho$ and $b_K$. The supplied line shape gives
$$
\mathcal R(E)\equiv\frac{\sigma_\rho}{\sigma_K}
=\frac{3b_\rho}{0.69b_K}\frac{\Gamma_\rho^2}{\Gamma_K^2}
\frac{(E^2-m_K^2)^2+E^2\Gamma_K^2}
{(E^2-m_\rho^2)^2+E^2\Gamma_\rho^2}.
$$
The distinction between <inclusive and exclusive resonance branching factors> matters here. For inclusive resonance formation, sum over all decays, so $b_\rho=b_K=1$. At $E=m_K$ the small <kaon> width cancels from the peak value:
$$
\boxed{\mathcal R_{\rm inclusive}(498)=\frac3{0.69}
\frac{498^2\,147^2}{(498^2-769^2)^2+498^2\,147^2}
=0.1890.}
$$
If instead the measured cross section is specifically resonant elastic $\pi^+\pi^-\to\pi^+\pi^-$, its exit <branching fractions> equal the entrance ones: $b_\rho=1$, $b_K=0.69$. Then
$$
\boxed{\mathcal R_{\pi\pi}(498)=0.1890/0.69=0.2740.}
$$
The source says production without specifying an observed decay channel, so both conventions are made explicit rather than silently squaring a <branching fraction> for an inclusive measurement. The quoted <kaon> width and two-pion <branching fraction> identify the short-lived propagation state $K_S$; a pure flavor state $K^0$ is a superposition of short- and long-lived states. This exercise uses the stipulated $K_S$ resonance model and equal omitted proportionality factors.
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