The magnetic field of an electromagnetic wave is given by $\overrightarrow{B}=1.6\times {10}^{-6}\cos \left(2\times {10}^{7}z+6\times {10}^{15}t\right)\left(2\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)\frac{\mathrm{Wb}}{{\mathrm{m}}^2}$. The associated electric field will be:

$$\begin{gathered} \frac{E_0}{B_0}=c \\ {E}_0=B_{0}c=\left(\sqrt{2^2+1^2}\right)\times 1.6\times {10}^{-6}\times 3\times {10}^8=\sqrt{5}\times 4.8\times {10}^2 V/m \end{gathered}$$

Direction of $\overrightarrow{B}$ is along $2\hat{\mathrm{i}}+\hat{\mathrm{j}}$.

$\Rightarrow \hat{B}=\frac{1}{\sqrt{5}}\left(2\hat{i}+\hat{j}\right)$

velocity is along $-z$ - direction.

$\Rightarrow \hat{c}=-\hat{k}$

so, $\hat{E}=\hat{B}\times \hat{c}=\frac{1}{\sqrt{5}}\left(2\hat{i}+\hat{j}\right)\times \left(-\hat{k}\right)=\frac{1}{\sqrt{5}}\left(-\hat{i}+2\hat{j}\right)$

$\overrightarrow{E}=E_0\hat{E}$