(i) The electromagnetic wave carry energy which is due to electric field vector and magnetic field vector. In electromagnetic wave, $E$ and $B$ vary from point to point and from moment to moment. Let $E$ and $B$ be their time averages.

The energy density due to electric field $E$ is

$$u_E=\frac{1}{2} \varepsilon_0 E^2$$

The energy density due to magnetic field $B$ is

$$u_B=\frac{1}{2} \frac{B^2}{\alpha_0}$$

Total average energy density of electromagnetic wave

$$u_{a v}=u_E+u_B=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} \frac{B^2}{\alpha_0}$$

Let the EM wave be propagating along $z$-direction. The electric field vector and magnetic field vector be represented by

$$\begin{aligned} & E=E_0 \sin (k z-\omega t) \\ & B=B_0 \sin (k z-\omega t) \end{aligned}$$

The time average value of $E^2$ over complete cycle $=\frac{E_0^2}{2}$

and time average value of $B^2$ over complete cycle $=\frac{B_0^2}{2}$

$$\begin{aligned} u_{\mathrm{av}} & =\frac{1}{2} \frac{\varepsilon_0 E_0^2}{2}+\frac{1}{2} \alpha_0\left(\frac{B_0^2}{2}\right) \\ & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{B_0^2}{4 \alpha_0} \end{aligned}$$

$$\begin{aligned} &\text { (ii) We know that } E_0=c B_0 \text { and } c=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}}\\ &\begin{aligned} & \therefore \quad \frac{1}{4} \frac{B_0^2}{\alpha_0}=\frac{1}{4} \frac{E_0^2 / C^2}{\alpha_0}=\frac{E_0^2}{4 \propto_0} \times \propto_0 \varepsilon_0=\frac{1}{4} \varepsilon_0 E_0^2 \\ & \therefore \quad u_B=u_E \end{aligned} \end{aligned}$$

$$\begin{aligned} \text{Hence,}\quad U_{\mathrm{av}} & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \frac{B_0^2}{\alpha_0} \\ & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \varepsilon_0 E_0^2 \\ & =\frac{1}{2} \varepsilon_0 E_0^2=\frac{1}{2} \frac{B_0^2}{\alpha_0} \end{aligned}$$

$$\begin{aligned} &\text { Time average intensity of the wave }\\ &I_{\mathrm{av}}=U_{\mathrm{av}} C=\frac{1}{2} \varepsilon_0 E_0^2 C=\frac{1}{2} \varepsilon_0 E_0^2 \end{aligned}$$