Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric field energy density is given by
$$u_E=\frac{1}{2}\epsilon_0 E^2\tag 1$$
and the magnetic field energy density is given by
$$u_B=\frac{1}{2\mu_0}B^2\tag 2$$
We also know, for the same electromagnetic wave, that
$$E=cB$$
Squaring both sides;
$$E^2=c^2B^2$$
Plug from (1) and (2),
$$\dfrac{\color{red}{\bf\not} 2u_E}{\epsilon_0}=c^2 (\color{red}{\bf\not} 2\mu_0u_B)$$
where $c^2=1/\mu_0 \epsilon_0$,
$$\dfrac{ u_E}{\color{red}{\bf\not} \epsilon_0}=\dfrac{1}{\color{red}{\bf\not} \mu_0 \color{red}{\bf\not} \epsilon_0} \color{red}{\bf\not} \mu_0u_B$$
$$\boxed{u_E=u_B}$$
$$\color{blue}{\bf [b]}$$
We know that the intensity of an electromagnetic wave is given by
$$I=\frac{1}{2}\epsilon_0cE^2\tag 3$$
The total energy density is given by
$$u_{\rm total}=u_E+u_B=2u_E=2u_B $$
$$u_{\rm total}=2u_E$$
Plug from (1),
$$u_{\rm total}= 2\left[\frac{1}{2}\epsilon_0 E^2\right] $$
$$u_{\rm total}= \epsilon_0 E^2$$
Plug $E^2$ from (3)
$$u_{\rm total}= \epsilon_0 \left[\dfrac{2I}{\epsilon_0c}\right] $$
$$u_{\rm total}= \left[\dfrac{2I}{ c}\right] $$
Plug the known;
$$u_{\rm total}= \dfrac{2(1000) }{ (3\times 10^8) }$$
$$u_{\rm total}=\color{red}{\bf 6.67\times 10^{-6}}\;\rm J/m^3$$
