As we know,

$$\begin{gathered} c=\frac{E_0}{B_0} \\ \Rightarrow E_0=B_{0}c \\ =\left(2\times {10}^{-8}\right)\left(3\times {10}^8\right) \\ =6 \mathrm{V} {\mathrm{m}}^{-1} \end{gathered}$$

Now, direction of propagation of wave

 $$\begin{gathered} =\hat{E}\times \hat{B} \\ \Rightarrow -\hat{\mathrm{j}}=\hat{E}\times \hat{\mathrm{k}} \\ \Rightarrow \hat{E}=\hat{\mathrm{i}} \end{gathered}$$

Angular frequency will remain the same. Therefore electric field can be written as,

$\overrightarrow{E}=\left(6\right)\cos \left[\pi \times {10}^{15}\left(t+\frac{y}{c}\right)\right] \hat{\mathrm{i}} \mathrm{V} {\mathrm{m}}^{-1}$

$$