= Solution
With $[\mathrm{H_2}]=5\times10^{-6}[\mathrm M]$, a reaction of an <electronically excited oxygen atom> with <hydrogen> gives one OH and, after rapid H conversion, one HO$_2$. The radical source becomes
$$
S'=\frac{2J_1[\mathrm O_3]\left(k_3[\mathrm{H_2O}]+k_{10}[\mathrm{H_2}]\right)}
{k_2[\mathrm M]+k_3[\mathrm{H_2O}]+k_{10}[\mathrm{H_2}]}
\simeq2.85\times10^4\,\mathrm{cm^{-3}s^{-1}}.
$$
This is about 4.75 times the original source. Holding <water>, <oxygen>, NO and radiation fixed keeps the fast radical partition approximately unchanged, while bimolecular termination makes total radical abundance scale as $\sqrt S$. Thus
$$
[\mathrm{HO_2}]'\simeq7.67\times10^8\,\mathrm{cm^{-3}},\qquad
\boxed{L'_{\rm HO_x}\simeq2.76\times10^6\,\mathrm{cm^{-3}s^{-1}}},
$$
\b[about 2.18 times the original catalytic loss]. This predicts a stronger ozone-loss tendency within the specified stationary box model, not a numerically determined percentage change in the <ozone> layer. The latter requires coupled odd-oxygen production, changing <ozone> <concentrations>, transport and the longer-term <water> budget; those have been held fixed here.
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