1. Average energy density in a plane EM wave

For an electromagnetic wave, the instantaneous total energy density is $u = u_E + u_B$ where $u_E = \frac{1}{2}\varepsilon_0 E^2, \qquad u_B = \frac{1}{2\mu_0}B^2$

In a plane EM wave, the electric and magnetic contributions are equal, so $u = \varepsilon_0 E^2$ for the instantaneous total energy density.

Since the electric field varies sinusoidally, $E = E_0 \sin \omega t$ we have $u = \varepsilon_0 E_0^2 \sin^2 \omega t$

Therefore, the average total energy density is $\langle u \rangle = \varepsilon_0 E_0^2 \langle \sin^2 \omega t \rangle$ Using $\langle \sin^2 \omega t \rangle = \frac{1}{2}$ we get $\langle u \rangle = \frac{1}{2}\varepsilon_0 E_0^2$

2. Substitute the given values

Given: $E_0 = 50\ \text{V m}^{-1}, \qquad \varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\text{/N m}^2$

So, $\langle u \rangle = \frac{1}{2}(8.85 \times 10^{-12})(50)^2$

3. Calculate step-by-step

First, $50^2 = 2500$

Then, $(8.85 \times 10^{-12})(2500) = 2.2125 \times 10^{-8}$

Now divide by 2: $\langle u \rangle = \frac{2.2125 \times 10^{-8}}{2} = 1.10625 \times 10^{-8}\ \text{J m}^{-3}$

4. Match with the options

$\boxed{\langle u \rangle = 1.106 \times 10^{-8}\ \text{J m}^{-3}}$

This matches Option D.

5. Note on frequency

The given frequency $5 \times 10^{10}\,\text{Hz}$ is not needed in this calculation, because average energy density depends only on the field amplitude $E_0$.

6. Comparison with stored correct answer

Stored correct answer: D

Derived answer: D

They agree.