Energy Density & Intensity question from JEE Main 2015

Answer: (d)

1. $I = \dfrac{P}{4\pi r^2} = \dfrac{0.1}{4\pi \times 1^2} \approx 7.96 \times 10^{-3}\ \text{W/m}^2$
2. $I = \tfrac{1}{2}\varepsilon_0 c E_0^2 \Rightarrow E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = \sqrt{\dfrac{2 \times 7.96 \times 10^{-3}}{2.655 \times 10^{-3}}}$
3. $E_0 = \sqrt{6.0} \approx 2.45\ \text{V/m}$, so the answer is (d).