Answer
$I=3.32\times 10^{-2}W/m^2$
Work Step by Step
The intensity of light is equal to the average value of the Poynting vector, which is $$S=\frac{E_mB_m}{2\mu_o}$$ Using the relation that $E_m=cB_m$ and $B_m=\frac{E_m}{c}$, the expression becomes $$S=\frac{E_m^2}{2c\mu_o}$$ Substituting the maximum value of $E_m=5.00V/m$ yields $$S=I=\frac{(5.00V/m)^2}{2(3.00\times 10^8m/s)(4\pi \times 10^{-7} T m)}=3.32\times 10^{-2}W/m^2$$
