= Solution
Convert the supplied concentrations to $[\mathrm{NO}]=1.00\times10^{-7}\,\mathrm M$ and $[\mathrm{O_2}]=3.00\times10^{-5}\,\mathrm M$. Then
$$
\boxed{[\mathrm{ONOO}]_{\rm ss}
=\frac{189(1.00\times10^{-7})(3.00\times10^{-5})}{6500+(1.0\times10^9)(1.00\times10^{-7})}
=8.59\times10^{-14}\,\mathrm M.}
$$
The combined loss coefficient is $6600\,\mathrm{s^{-1}}$. Neglecting the smaller NO-dependent loss would give $8.72\times10^{-14}\,\mathrm M$, a close low-NO approximation, but the full expression is just as easy to evaluate.
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