(i) Due to electric field vector and magnetic field vector the electromagnetic wave carry energy. In electromagnetic wave, $\mathrm{E}$ and $\mathrm{B}$ vary with time from point to point and from moment to moment.
Consider $E$ and $B$ be the time averages.
So, the energy density due to electric field $\mathrm{E}$ is
$$
\mathrm{u}_{\mathrm{E}}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2
$$
And the energy density due to magnetic field $\mathrm{B}$ is
$$
\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \frac{\mathrm{B}^2}{\mu_0}
$$
So, total average energy density of electromagnetic wave
$$
\mathrm{u}_{\mathrm{av}}=\mathrm{u}_{\mathrm{E}}+\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2+\frac{1}{2} \frac{\mathrm{B}^2}{\mu_0}
$$
Consider the electromagnetic wave propagating along $\mathrm{z}-$ direction. The electric field vector and magnetic field vector are :
$$
\begin{aligned}
&\mathrm{E}=\mathrm{E}_0 \sin (k z-\omega \mathrm{t}) \\
&\mathrm{B}=\mathrm{B}_0 \sin (k z-\omega \mathrm{t})
\end{aligned}
$$
Then thetimeaveragevalue of $\mathrm{E}^2$ over complete cycle= $\frac{\mathrm{E}_0^2}{2}$
Similarly the time average value of $\mathrm{B}^2$ over complete cycle $=\frac{\mathrm{B}_0^2}{2}$
So, $\mathrm{u}_{\mathrm{av}}=\frac{1}{2} \frac{\varepsilon_0 \mathrm{E}_0^2}{2}+\frac{1}{2} \mu_0\left(\frac{\mathrm{B}_0^2}{2}\right)$
Hence, $\quad \mathrm{u}_{\mathrm{av}}=\frac{1}{4}\left[\varepsilon_0 \mathrm{E}_0^2+\frac{\mathrm{B}_0^2}{4 \mu_0}\right]$
(ii) As we know that the relationship between $\varepsilon_0$ and $\mathrm{E}_0$ is $\mathrm{E}_0=\mathrm{cB}_0$ and $\mathrm{c}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
And, From (i) part, when, $u_{a v}=0$
$$
\frac{1}{4} \frac{B_0^2}{\mu_0}=\frac{-1}{4} \varepsilon_0 E_0^2=\frac{-1}{4} \frac{E_0^2}{\mu_0 \mathrm{c}^2}
$$
when (-ve) neglected.
So, $\frac{1}{4} \frac{\mathrm{B}_0^2}{\mu_0}=\frac{1}{4} \frac{\mathrm{E}_0^2 / \mathrm{c}^2}{\mu_0}=\frac{\mathrm{E}_0^2}{4 \mu_0} \times \mu_0 \varepsilon_0$
$$
\frac{1}{4} \frac{\mathrm{B}_0^2}{\mu_0}=\frac{1}{4} \frac{\mathrm{E}_0^2}{\varepsilon_0}
$$
or $\quad \frac{1}{2} \frac{\mathrm{B}_0^2}{\mu_0}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2$
$\therefore \quad \mathrm{u}_{\mathrm{B}}=\mathrm{u}_{\mathrm{E}}$
Again from (i) part,
$$
\begin{aligned}
\mathrm{u}_{\mathrm{av}} &=\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2+\frac{1}{4} \frac{\mathrm{B}_0^2}{\mu_0}\left[\because \mathrm{c}=\frac{\mathrm{E}_0}{\mathrm{~B}_0} \Rightarrow \mathrm{c}^2=\left(\frac{\mathrm{E}_0^2}{\mathrm{~B}_0^2}\right)\right] \\
&=\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2+\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2 \\
\mathrm{u}_{\mathrm{av}} &=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2=\frac{1}{2} \frac{\mathrm{B}_0^2}{\mu_0} \quad\left[\because \mathrm{E}_0^2=\frac{\mathrm{B}_0}{\mu_0 \varepsilon_0}\right] \\
\therefore & \mathrm{u}_{\mathrm{E}}=\mathrm{u}_{\mathrm{B}}
\end{aligned}
$$
So, time average intensity of the wave
$$
\mathrm{I}_{\mathrm{av}}=\mathrm{u}_{\mathrm{av}} \mathrm{C}=\frac{1}{2} \varepsilon_0 \mathrm{~B}_0^2 \mathrm{c}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{c}
$$
Hence, $\quad \mathrm{I}_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{c}$