= Solution
In a vertical-beam, effective-wavelength <Beer-Lambert law> approximation, the <optical depth> above altitude $z$ is $\tau(z)=\sigma_{\mathrm O_2}N_{\mathrm O_2}(z)+\sigma_{\mathrm O_3}N_{\mathrm O_3}(z)$. Using centimetres consistently gives
$$
\tau_{20}=1.50+12.60=14.10,\qquad
\tau_{40}=0.084+0.340=0.424.
$$
The transmitted <photon> flux is proportional to $e^{-\tau}$. If the parent-gas <absorption cross section> and <photodissociation> yield are otherwise the same, they cancel in the ratio of <photolysis rate coefficients>:
$$
\boxed{\frac{J(40\ \mathrm{km})}{J(20\ \mathrm{km})}=e^{\tau_{20}-\tau_{40}}
=e^{13.676}=8.70\times10^5.}
$$
The corresponding fractions of the unattenuated beam are about $7.52\times10^{-7}$ and $0.654$. Strong shielding by the overlying <oxygen> and especially <ozone> therefore makes destruction vastly faster higher in the <stratosphere>, even though the molecular absorption properties have not changed.
This numerical ratio uses the intended simplified radiation model. A real <photolysis rate coefficient> integrates wavelength-dependent <actinic flux>, <absorption cross section> and <quantum yield>. A nonvertical direct beam would use slant <optical depth> $\tau/\cos\theta$; diffuse scattering and the true solar spectrum would also matter. The supplied effective cross sections do not specify a unique broadband radiative-transfer calculation.
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