= Solution
The two initial reservoirs each contain one <chlorine> atom, so complete activation supplies a total <chlorine> <mixing ratio> of $2.2\times10^{-9}$. Under the specified approximation that active <chlorine> in sunlight is predominantly <chlorine monoxide>,
$$
[\mathrm{ClO}]=(2.2\times10^{-9})(1.2\times10^{18})
=2.64\times10^9\ \mathrm{molecules\,cm^{-3}}.
$$
The daytime loss is consequently
$$
L_d=1.4(9\times10^{-32})(1.2\times10^{18})(2.64\times10^9)^2
=\boxed{1.054\times10^6\ \mathrm{molecules\,cm^{-3}\,s^{-1}}.}
$$
For twelve sunlit hours per day, the daily <ozone> decrement is
$$
\Delta x_{\mathrm O_3}=\frac{L_d(43200)}{1.2\times10^{18}}
=\boxed{3.794\times10^{-8}\ \mathrm{day^{-1}}
=37.94\ \mathrm{ppb\,day^{-1}}=0.03794\ \mathrm{ppm\,day^{-1}}.}
$$
The full-day mean <concentration> loss is $5.269\times10^5\ \mathrm{molecules\,cm^{-3}\,s^{-1}}$. Maintaining that rate would consume the initial <ozone> in about $79.1$ days. This is a fixed-active-chlorine estimate: reservoir reformation, transport and <chlorine> partitioning change during a real depletion event, and the approximation fails when insufficient <ozone> remains to recycle <chlorine> rapidly.
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