If Electric field intensity of a uniform plane electro magnetic...

The relation between magnetic field and electric field is given by,

$c = \dfrac{{E}_{0}}{{B}_{0}} ⇒ {B}_{0} = \dfrac{{E}_{0}}{c}$

And magnetic intensity is given by,

${H}_{0} = \dfrac{{B}_{0}}{{\mu}_{0}} ⇒ {H}_{0} = \dfrac{{E}_{0}}{c{\mu}_{0}}$

Also, $\overset{\to}{E} and$\overset{\to}{B}$are perpendicular to each other.</p><p>Therefore,</p><p>\left|{H}_{y0}\right| = \dfrac{301 . 6}{3 \times {10}^{8} \times 4\pi \times {10}^{- 7}} = 0 . 8A{m}^{- 1}\left|{H}_{x0}\right| = \dfrac{452 . 4}{3 \times {10}^{8} \times 4\pi \times {10}^{- 7}} = 1 . 2A{m}^{- 1}$

Hence, C is the answer.