Ans:(a),(b) and (c).

Explanation:

The direction of electromagnetic wave is always along the direction of vector product $\overrightarrow{\mathrm{E}}\times \overrightarrow{\mathrm{B}}$

from figure

$$\begin{gathered} \overrightarrow{\mathrm{E}}=\mathrm{E}\hat{\mathrm{i}} \\ \overrightarrow{\mathrm{B}}=\mathrm{B}\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{C}}=\mathrm{C}\hat{\mathrm{k}} \\ \therefore \overrightarrow{\mathrm{B}}=\mathrm{B}\hat{\mathrm{j}} \\ =\mathrm{B}\left( \hat{\mathrm{k}}\times \hat{\mathrm{i}} \right) \left[\because \hat{\mathrm{k}}\times \hat{\mathrm{i}}=\hat{\mathrm{j}}\right] \\ =\frac{\mathrm{E}}{\mathrm{C}}\left( \hat{\mathrm{k}}\times \hat{\mathrm{i}} \right) \left[\because \mathrm{C}=\frac{\mathrm{E}}{\mathrm{B}}\right] \\ =\frac{1}{\text{C}}\left(\hat{\mathrm{k}}\times \mathrm{E}\widehat{\mathrm{i}}\right) \\ \boxed{\overrightarrow{\mathrm{B}}=\frac{1}{\mathrm{C}}\left(\hat{\mathrm{k}}\times \hat{\mathrm{E}}\right)} \mathrm{which} \mathrm{verifies} \mathrm{option} \left(\mathrm{a}\right) \\ \left(\mathrm{b}\right) \overrightarrow{\mathrm{E}}=\mathrm{E}\hat{\mathrm{i}} \\ =\mathrm{CB}\left(\hat{\mathrm{i}}\right) \left[\because \mathrm{C}=\frac{\mathrm{E}}{\mathrm{B}}\right] \\ =\mathrm{CB}\left(\hat{\mathrm{j}}\times \hat{\mathrm{k}}\right)=\mathrm{C}\left(\mathrm{B}\hat{\mathrm{j}}\times \hat{\mathrm{k}}\right)=\mathrm{C}\left(\overrightarrow{\mathrm{B}}\times \hat{\mathrm{k}}\right) \mathrm{which} \mathrm{verifies} \mathrm{option} \left(\mathrm{b}\right) \\ \left(\mathrm{c}\right) \hat{\mathrm{k}}.\overrightarrow{\mathrm{E}}=\hat{\mathrm{k}}.\mathrm{E}\hat{\mathrm{i}} =\mathrm{E}\left(\hat{\mathrm{k}}.\hat{\mathrm{i}}\right)=0 \mathrm{and} \\ \hat{\mathrm{k}}.\overrightarrow{\mathrm{B}}=\hat{\mathrm{k}}.\mathrm{B}\hat{\mathrm{i}} =\mathrm{B}\left(\hat{\mathrm{k}}.\hat{\mathrm{i}}\right)=0 \\ \left(\mathrm{d}\right) \hat{\mathrm{k}}.\overrightarrow{\mathrm{E}}=\hat{\mathrm{k}}.\mathrm{E}\hat{\mathrm{i}} =\mathrm{E}\left(\hat{\mathrm{k}}.\hat{\mathrm{i}}\right)=\mathrm{E}\hat{\mathrm{j}} \mathrm{and} \\ \hat{\mathrm{k}}.\overrightarrow{\mathrm{B}}=\hat{\mathrm{k}}.\mathrm{B}\hat{\mathrm{i}} =\mathrm{B}\left(\hat{\mathrm{k}}.\hat{\mathrm{i}}\right)=-\mathrm{B}\hat{\mathrm{i}} \end{gathered}$$