= Solution
The quoted angular diameter gives half-angle $\theta=0.005$. At distance $r=10^{13}\,\mathrm m$, the uniform beam covers area $A\simeq\pi(r\theta)^2=7.85\times10^{21}\,\mathrm{m^2}$. Its intensity is $I=P/A\simeq2.93\times10^{-21}\,\mathrm{W\,m^{-2}}$. For a vacuum electromagnetic wave, $I=\epsilon_0cE_{\rm rms}^2$, so
$$
\boxed{E_{\rm rms}\simeq1.05\times10^{-9}\,\mathrm{V\,m^{-1}}.}
$$
No extra factor two belongs in this rms relation. The <frequency> fixes the wavelength but is unnecessary once the total power and stipulated beam divergence are specified. This is the ideal directed-beam estimate, before propagation or receiver losses.
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