The intensity of TV broad cast station of $E=800\sin ({10}^{9}t-kx)V/M$ is…. and the wave length in meter is….

$I=\frac{1}{2}{\epsilon }_0{E}^{2}C$

$\lambda =\frac{C}{n}$

$\begin{array}{r}\mathrm{E}=800\sin \left({10}^9\mathrm{t}-\mathrm{kx}\right){\mathrm{Vm}}^{-1}\\ \mathrm{I}={\mathrm{u}}_{\text{avg }}\times \mathrm{C}\\ =\frac{1}{2}{\epsilon }_0{\mathrm{E}}_0^2\times \mathrm{C}\\ =\frac{1}{2}\times 8.85\times {10}^{-12}\times (800{)}^2\times 3\times {10}^8\\ =849.6\\ \approx 850{\mathrm{Wm}}^{-2}\end{array}$

$\begin{array}{r}\omega ={10}^9,2\pi \mathrm{n}={10}^9,\mathrm{n}=\frac{{10}^9}{2\pi };\mathrm{C}=\mathrm{n}\lambda \\ \lambda =\frac{\mathrm{C}}{\mathrm{n}}=\frac{3\times {10}^8}{{10}^9}\times 2\pi ;\\ \lambda =0.6\pi \mathrm{m}\end{array}$