The average value of electric energy density in an electromagnetic wave is. [where $E_0$ is peak value]

An electromagnetic wave of frequency $1 \times 10^{14} \,Hz$ is propagating along the $z$-axis. The amplitude of the electric field is $4 \,Vm^{-1}$. What is the energy density of the electric field? (Permittivity of free space $\varepsilon_0 = 8.8 \times 10^{-12} \,C^2 \,N^{-1} \,m^{-2}$)

Given below are two statements:
Statement $I$: Electromagnetic waves are not deflected by electric and magnetic fields.
Statement $II$: The amplitude of the electric field and the magnetic field in electromagnetic waves are related to each other as $E_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} B_0$.
In the light of the above statements,choose the correct answer from the options given below:

The magnetic field of a plane electromagnetic wave is $\overrightarrow{B} = 3 \times 10^{-8} \sin [200 \pi(y + ct)] \hat{i} \, T$,where $c = 3 \times 10^{8} \, m/s$ is the speed of light. The corresponding electric field is:

$A$ plane electromagnetic wave is moving in free space with velocity $c = 3 \times 10^8 \ m/s$ and its electric field is given as $\vec{E} = 54 \sin(kz - \omega t) \hat{j} \ V/m$, where $\hat{j}$ is the unit vector along the $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:

For an electromagnetic wave propagating through vacuum, $\vec{k}$, $\vec{E}$, and $\omega$ represent the propagation vector, electric field, and angular frequency, respectively. The magnetic field associated with this wave is represented by: