A plane wave in vacuum has a peak electric field of 450 V/m…

Step 1. The energy density. $\bar{u} = \tfrac{1}{2}\varepsilon_0 E_0^2 = \tfrac{1}{2} \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times (450\ \text{V/m})^2 = 0.896\ \mu\text{J/m}^3$.

Step 2. The share of the magnetic field. With $E = cB$ and $c^2 = 1/\mu_0\varepsilon_0$, the magnetic energy density $B^2/2\mu_0$ equals $\tfrac{1}{2}\varepsilon_0E^2$. So the magnetic field holds $50\ \%$ of the energy.

Step 3. The intensity. $I = \bar{u}c = 0.896 \times 10^{-6}\ \text{J/m}^3 \times 3.00 \times 10^{8}\ \text{m/s} = 269\ \text{W/m}^2$.