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

We have the magnetic field of the plane electromagnetic wave in vacuum written as

$$\vec{B}(z,t)=1.6\times10^{-6}\;\cos\!\left(2\times10^{7}\,z+6\times10^{15}\,t\right)\,(2\hat{i}+\hat{j})\;\;\frac{\text{Wb}}{\text{m}^2}.$$

The argument of the cosine is $$kz+\omega t,$$ where $$k=2\times10^{7}\,\text{m}^{-1}$$ and $$\omega=6\times10^{15}\,\text{s}^{-1}.$$ Because the sign in front of $$\omega t$$ is positive, the phase travels in the negative $$z$$-direction. Hence the unit vector in the direction of propagation is

$$\hat{n}=-\hat{k}=-\hat{z}.$$

For a plane electromagnetic wave in free space we know two standard relations:

1. Magnitude relation $$E_0=c\,B_0,$$ where $$c=3\times10^{8}\;\text{m/s}.$$

2. Vector relation $$\vec{B}=\dfrac{1}{c}\,\hat{n}\times\vec{E}.$$

From the first relation we can calculate the amplitude of the electric field. The magnetic amplitude is

$$B_0=1.6\times10^{-6}\;\text{Wb/m}^2.$$

So, using $$E_0=cB_0,$$ we get

$$E_0=(3\times10^{8})\,(1.6\times10^{-6})=4.8\times10^{2}\;\text{V/m}.$$

Now we determine the direction of $$\vec{E}$$. From $$\vec{B}=\dfrac{1}{c}\,\hat{n}\times\vec{E}$$ we rearrange to obtain

$$\vec{E}=-c\,\hat{n}\times\vec{B}.$$

Substituting $$\hat{n}=-\hat{z}$$ gives

$$\vec{E}= -c\,(-\hat{z})\times\vec{B}=c\,\vec{B}\times\hat{z}.$$

We therefore need the cross-product $$\vec{B}_\text{dir}\times\hat{z},$$ where $$\vec{B}_\text{dir}=2\hat{i}+\hat{j}$$ is the direction part of the magnetic field.

Writing the cross product explicitly,

$$\vec{B}_\text{dir}\times\hat{z}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\[4pt] 2&1&0\\[4pt] 0&0&1 \end{vmatrix} = \hat{i}(1\cdot1-0\cdot0)-\hat{j}(2\cdot1-0\cdot0)+\hat{k}(2\cdot0-1\cdot0) = \hat{i}-2\hat{j}+0\hat{k}.$$

So,

$$\vec{B}\times\hat{z}=(\hat{i}-2\hat{j}).$$

Multiplying by the amplitude $$B_0$$ and by $$c,$$ the full electric field becomes

$$\vec{E}(z,t)=c\,B_0\;\cos\!\left(2\times10^{7}\,z+6\times10^{15}\,t\right)\,(-\hat{i}+2\hat{j})$$ $$=4.8\times10^{2}\;\cos\!\left(2\times10^{7}\,z+6\times10^{15}\,t\right)\;(-\hat{i}+2\hat{j})\;\frac{\text{V}}{\text{m}}.$$

This expression matches exactly with Option C.

Hence, the correct answer is Option C.