1. Given electric field

The wave is $E = (50\,\text{N C}^{-1})\sin\omega\left(t-\frac{x}{c}\right)$ So the amplitude of electric field is $E_0 = 50\,\text{N C}^{-1}$

2. Energy density of an electromagnetic wave

The instantaneous total energy density in an electromagnetic wave is $u = u_E + u_B = \epsilon_0 E^2$ because electric and magnetic parts are equal, and $u_E = \frac{1}{2}\epsilon_0 E^2, \qquad u_B = \frac{1}{2}\epsilon_0 E^2$

For a sinusoidal wave, the average value of $E^2$ is $\langle E^2 \rangle = \frac{E_0^2}{2}$ Hence average energy density is $u_{\text{avg}} = \epsilon_0 \langle E^2 \rangle = \frac{1}{2}\epsilon_0 E_0^2$

3. Substitute values

Given $\epsilon_0 = 8.8\times 10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}, \qquad E_0=50$

So, $u_{\text{avg}} = \frac{1}{2}(8.8\times 10^{-12})(50)^2$ $= \frac{1}{2}(8.8\times 10^{-12})(2500)$ $= 4.4\times 10^{-12}\times 2500$ $= 1.1\times 10^{-8}\,\text{J m}^{-3}$

4. Use total energy = energy density $\times$ volume

Given total energy contained in the cylinder is $U = 5.5\times 10^{-12}\,\text{J}$

Thus, $V = \frac{U}{u_{\text{avg}}}$ $= \frac{5.5\times 10^{-12}}{1.1\times 10^{-8}}$ $= 5\times 10^{-4}\,\text{m}^3$

5. Convert to cm$^3$

Since $1\,\text{m}^3 = 10^6\,\text{cm}^3$ therefore, $V = 5\times 10^{-4}\times 10^6 = 5\times 10^2 = 500\,\text{cm}^3$

6. Final answer

$\boxed{500}$

The derived answer matches the stored correct answer.