= Solution
The initial <ozone> <concentration> is $(3\times10^{-6})[M]=3.60\times10^{12}\ \mathrm{molecules\,cm^{-3}}$. Over fifty days with twelve active hours per day, the required combined daytime rate is
$$
L_{\rm req}=\frac{3.60\times10^{12}}{50(43200)}
=1.667\times10^6\ \mathrm{molecules\,cm^{-3}\,s^{-1}}.
$$
The <mixed chlorine-bromine ozone-loss cycle> also consumes two <ozone> <molecules> per radical-pair reaction, so $L_{\rm Br}=2k_6[\mathrm{ClO}][\mathrm{BrO}]$. Subtracting the dimer contribution yields
$$
[\mathrm{BrO}]=\frac{L_{\rm req}-L_d}{2k_6[\mathrm{ClO}]}
=1.935\times10^7\ \mathrm{molecules\,cm^{-3}},
\qquad
\boxed{x_{\rm BrO}=1.612\times10^{-11}=16.1\ \mathrm{ppt}.}
$$
The same result follows by assigning the dimer $1.897$ ppm of the fifty-day loss and the mixed cycle the remaining $1.103$ ppm. This uses the same daylight fraction for both cycles and retains the earlier fixed-ClO approximation. It estimates a sunlit <BrO> abundance, not the total inventory of <bromine> in all reservoirs.
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