Answer
$4.50\times10^{-6}J$.
Work Step by Step
Use equation 22–6a to link the instantaneous value of energy density to the electric field strength.
$$u=\epsilon_oE^2$$
Now use equation 22–7 to relate this to the intensity of sunlight.
$$\overline{u}=\frac{\overline{I}}{c}=\frac{1}{2}\epsilon_oE^2$$
Now find the average energy contained in a volume. Use the intensity value from Example 22-4.
$$\overline{U}=\overline{u}V=\frac{\overline{I}}{c}V$$
$$\overline{U} =\frac{1350W/m^2}{3.00\times10^8m/s}(1.00m^3)= 4.50\times10^{-6}J $$