A monochromatic beam of light has a frequency $\nu =\frac{3}{2\mathrm{\pi }}\times {10}^{12} \mathrm{Hz}$ and is propagating along the direction $\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$. It is polarized along the $\hat{\mathrm{k}}$ direction. The acceptable form for the magnetic field is:

$\overrightarrow{\mathrm{E}}\times \overrightarrow{\mathrm{B}}$ gives the direction of wave propagation.

$\Rightarrow \left(\hat{\mathrm{k}}\times \overrightarrow{\mathrm{B}}\right) \parallel \left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$

Let the $\mathrm{a}\hat{\mathrm{i}} + \mathrm{b}\hat{\mathrm{j}}$ be the unit vector along $\overrightarrow{\mathrm{B}.}$

then: 

$$\begin{gathered} \hat{\mathrm{k}} \times (\mathrm{a}\hat{\mathrm{i}} + \mathrm{b}\hat{\mathrm{j}}) = \frac{\left(\hat{\mathrm{i}} +\hat{\mathrm{j}} \right)}{\sqrt{2}} \\ \Rightarrow \mathrm{a}\hat{\mathrm{i}} + \mathrm{b}\widehat{\mathrm{j} } = \frac{\left(\hat{\mathrm{i}} - \hat{\mathrm{j}} \right)}{\sqrt{2}} \end{gathered}$$

Hence, direction of the magnetic field is along $\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}}.$

The magnitude of amplitude of magnetic field is ${\mathrm{B}}_0 = \frac{{\mathrm{E}}_0}{\mathrm{c}}$.

$\mathrm{K}=\frac{2\mathrm{\pi \nu }}{\mathrm{c}}=\frac{3\times {10}^{12}}{3\times {10}^8}={10}^4$

Therefore, $\overrightarrow{\mathrm{B}}=\frac{{\mathrm{E}}_0}{\mathrm{C}}\left(\frac{\hat{\mathrm{i}} - \hat{\mathrm{j}}}{\sqrt{2}}\right)\cos \left[{10}^4\left(\frac{\hat{\mathrm{i}} + \hat{\mathrm{j}}}{\sqrt{2}}\right)·\overrightarrow{\mathrm{r}}-\left(3\times {10}^{12}\right)\mathrm{t}\right]$