Wave speed and refractive index. Read off $\omega =2\pi (5\times 10^{14})$ and $k=2\pi \cdot 10^7/3$. The phase speed in the medium is

$v=\frac{\omega }{k}=\frac{5\times 10^{14}}{10^7/3}=1.5\times 10^8{\text{m s}}^{-1}$.

So the refractive index is $n=c/v=(3\times 10^8)/(1.5\times 10^8)=2$. Option (D) is correct.

Magnetic field amplitude. The peak $E$-field amplitude is $|{\vec{E}}_0|=30\sqrt{2^2+1^2}=30\sqrt{5}{\text{V m}}^{-1}$, so

$|{\vec{B}}_0|=\frac{|{\vec{E}}_0|}{v}=\frac{30\sqrt{5}}{1.5\times 10^8}=2\sqrt{5}\times 10^{-7}\text{T}$.

Direction of $\vec{B}$. Propagation is along $+\hat{z}$, so $\hat{B}$ is along $\hat{k}\times \hat{E}$. With $\hat{E}=(2\hat{x}+\hat{y})/\sqrt{5}$,

$\hat{k}\times \hat{E}=\hat{z}\times \frac{2\hat{x}+\hat{y}}{\sqrt{5}}=\frac{2\hat{y}-\hat{x}}{\sqrt{5}}$.

Therefore $\vec{B}=2\sqrt{5}\times 10^{-7}\cdot \frac{-\hat{x}+2\hat{y}}{\sqrt{5}}\sin [\cdot ]=10^{-7}(-2\hat{x}+4\hat{y})\sin [\cdot ]$.

Hence $B_x=-2\times 10^{-7}\sin [\cdot ]{\text{Wb m}}^{-2}$ (option A correct) and $B_y=4\times 10^{-7}\sin [\cdot ]$, not $2\times 10^{-7}$, so option (B) is wrong.

Polarization angle. The electric vector lies along $(2,1)$ in the $xy$-plane, so the angle with the $x$-axis is ${\tan }^{-1}(1/2)\approx 26.57^{\circ }$, not $30^{\circ }$. Option (C) is wrong.

Correct options: (A) and (D).