= Solution
Approximate the beam as uniform over its circular cross section. Its mean <intensity> is $I=P/[\pi(D/2)^2]=1.27\times10^6\,\mathrm{W\,m^{-2}}$. For the <root mean square magnetic field from beam power> in a vacuum electromagnetic wave, $I=cB_{\rm rms}^2/\mu_0$, so
$$
\boxed{B_{\rm rms}=\sqrt{\frac{\mu_0P}{c\pi(D/2)^2}}=7.3\times10^{-5}\,\mathrm T.}
$$
This is an rms field, not the peak sinusoidal <amplitude>. A Gaussian-beam peak would need a stated diameter convention.
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