$$ B_0=\frac{E_0}{C}=\frac{60}{3 \times 10^8}=2 \times 10^{-7} T $$ $\hat{\mathrm{E}} \times \hat{\mathrm{B}}$ must be direction of propagation. So, $\hat{\mathrm{B}} \rightarrow \mathrm{z}$-axis $$ \begin{aligned} & k=\frac{2 \pi}{\lambda}=\frac{\pi}{4} \times 10^3 \mathrm{~m}^{-1} \\ & E_y=60 \sin \left[\frac{\pi}{4} \times 10^3\left(\mathrm{x}-3 \times 10^8 \mathrm{t}\right)\right] \hat{j} V_m^{-1} \\ & E_z=2 \times 10^{-7} \sin \left[\frac{\pi}{4} \times 10^3\left(\mathrm{x}-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{kT}} \end{aligned} $$