Answer
Choice B.
Work Step by Step
From the information given earlier the electric field limit limits the intensity.
$$ I= \frac{1}{2}\epsilon_0 cE^2_{max}$$
$$ I_{max,E}= \frac{1}{2}\epsilon_0 c(\frac{350}{f})^2$$
Now express the intensity in terms of the magnetic field.
$$ I= \frac{1}{2}\epsilon_0 cE^2_{max}$$
$$ I_{max,B}= \frac{1}{2}\epsilon_0 c(c^2)B^2_{max}$$
$$ I_{max,B}= \frac{1}{2}\epsilon_0 c^3(\frac{5}{f})^2$$
Take the ratio of the two intensities.
$$\frac{ I_{max,E}}{ I_{max,B}}=\frac{1}{c^2}\frac{350^2}{5^2}=5.4\times10^{-14}$$
The allowed intensity using the electric field guideline is much smaller than the allowed intensity using the magnetic field guideline.
