The electric field component of monochromatic radiation is given by $\overrightarrow{\mathrm{E}}=2{\mathrm{E}}_0\hat{\mathrm{i}} \cos \left(\mathrm{kz}\right) \cos \left(\mathrm{\omega t}\right)$. Its magnetic field is then given by:

As, $\frac{\mathrm{dE}}{\mathrm{dz}}=-\frac{\mathrm{dB}}{\mathrm{dt}}$

$$\begin{gathered} \frac{\mathrm{dE}}{\mathrm{dz}}=-2{\mathrm{E}}_0\mathrm{ksinkz} \mathrm{cos\omega t}=-\frac{\mathrm{dB}}{\mathrm{dt}} \\ \Rightarrow \mathrm{dB}=+2{\mathrm{E}}_0\mathrm{k} \sin \left(\mathrm{kz}\right) \cos \left(\mathrm{\omega t}\right) \mathrm{dt} \end{gathered}$$

or, $\mathrm{B}=+2{\mathrm{E}}_{0}k \sin \left(\mathrm{kz}\right) \int \cos \left(\mathrm{\omega t}\right) \mathrm{dt}=+2{\mathrm{E}}_0\frac{\mathrm{k}}{\mathrm{\omega }}\sin \left(\mathrm{kz}\right) \sin \left(\mathrm{\omega t}\right)$

So, 

$$\begin{gathered} \frac{2{\mathrm{E}}_0}{2{\mathrm{B}}_0}=\frac{\mathrm{\omega }}{\mathrm{k}}=\mathrm{c} \\ \mathrm{B}=\frac{2{\mathrm{E}}_0}{\mathrm{c}}\sin \left(\mathrm{kz}\right) \sin \left(\mathrm{\omega t}\right) \\ \therefore \overrightarrow{\mathrm{B}}=\frac{2{\mathrm{E}}_0}{\mathrm{c}} \hat{\mathrm{j}} \sin \left(\mathrm{kz}\right) \sin \left(\mathrm{\omega t}\right) \end{gathered}$$