= Solution
The far fields obey $E_\theta=cB_\phi$. The radial component of their <Poynting vector> is $E_\theta B_\phi/\mu_0$. Averaging $\ddot p^2=p_0^2\omega^4\cos^2\omega(t-r/c)$ and integrating $\int\sin^2\theta\,d\Omega=8\pi/3$ gives
$$
\boxed{P=\frac{\mu_0p_0^2\omega^4}{12\pi c}.}
$$
For a small isotropically polarizable particle, $p_0=\alpha E_0$ and incident <intensity> is $I=\epsilon_0cE_0^2/2$. Thus the total <Rayleigh scattering> cross-section is
$$
\boxed{\sigma=\frac PI=\frac{\alpha^2\omega^4}{6\pi\epsilon_0^2c^4}
=\frac{8\pi^3\alpha^2}{3\epsilon_0^2\lambda^4}.}
$$
Each incident <linear polarization> gives the same integrated <power>; averaging <unpolarized light> does not add another factor of two. The scalar real <polarizability> approximation neglects absorption and molecular anisotropy.
For <Rayleigh attenuation from refractivity> in dilute air, <polarization> <density> is $\mathbf P=N\alpha\mathbf E$. Consequently $\epsilon_r=1+N\alpha/\epsilon_0$ and, with $\mu_r\simeq1$ and $n-1\ll1$,
$$
\boxed{n-1\simeq\frac{N\alpha}{2\epsilon_0}.}
$$
The ideal-gas <number density> at the supplied pressure and <temperature> is $N=P/(k_BT)=2.47\times10^{25}\,\mathrm{m^{-3}}$, yielding $\alpha=2.15\times10^{-40}\,\mathrm{C\,m^2\,V^{-1}}$. For an explicit representative visible <wavelength> $\lambda=550\,\mathrm{nm}$,
$$
\sigma\simeq5.33\times10^{-31}\,\mathrm{m^2},\qquad
\boxed{L_{\rm att}=\frac1{N\sigma}\simeq7.6\times10^4\,\mathrm m\approx76\,\mathrm{km}.}
$$
Direct-beam <intensity> follows $I(x)=I(0)e^{-x/L_{\rm att}}$. Equivalently $L_{\rm att}=3N\lambda^4/[32\pi^3(n-1)^2]$. Its strong <wavelength> dependence gives roughly $21$–$200\,\mathrm{km}$ across $400$–$700\,\mathrm{nm}$ if refractivity is held fixed. This is a uniform-density dry-air scattering estimate, not absorption or transmission through a vertically stratified atmosphere; aerosols and molecular depolarization corrections are outside the given model.
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