= Solution
Using an isothermal atmosphere with the stated scale height gives $p(30\,\mathrm{km})=10^5e^{-30/7}\,\mathrm{Pa}$ and $[\mathrm M]=p/(k_BT)=3.46\times10^{17}\,\mathrm{cm^{-3}}$ at $288\,\mathrm K$. The <water> <atmospheric mixing ratio> then gives $[\mathrm{H_2O}]=6.92\times10^{11}\,\mathrm{cm^{-3}}$. Applying steady state to the short-lived <electronically excited oxygen atom>,
$$
[\mathrm O({}^1D)]=\frac{J_1[\mathrm O_3]}{k_2[\mathrm M]+k_3[\mathrm{H_2O}]}
=21.7\,\mathrm{cm^{-3}}.
$$
Its <HOx family> source is $S=2k_3[\mathrm O({}^1D)][\mathrm{H_2O}]\simeq6.00\times10^3\,\mathrm{cm^{-3}s^{-1}}$. Balancing this with termination gives $2k_4[\mathrm{OH}][\mathrm{HO_2}]=S$. Combining with the rapid partition, or solving both stationary radical equations including termination, gives
$$
\boxed{[\mathrm{OH}]\simeq8.53\times10^4,\qquad
[\mathrm{HO_2}]\simeq3.52\times10^8\ \mathrm{cm^{-3}}}.
$$
Consequently the main catalytic odd-oxygen loss is
$$
\boxed{L_{\rm HO_x}\simeq1.27\times10^6\,\mathrm{cm^{-3}s^{-1}}}.
$$
The <Chapman cycle> loss is $L_{\rm Ch}=2k_9[\mathrm O][\mathrm O_3]=5.40\times10^5\,\mathrm{cm^{-3}s^{-1}}$, so the given HO$_x$ cycle is \b[about 2.35 times larger]. Both rates count two odd-oxygen units per completed O-plus-O$_3$ destruction event. The omitted $v_4\simeq3.00\times10^3$ correction is only about $0.24\%$ of the main HO$_x$ loss. The family abundance divided by its source is about $5.87\times10^4\,\mathrm s$, explaining why its absolute stationary state is a stronger assumption than rapid OH/HO$_2$ interconversion.
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