Average Energy Density

The total instantaneous energy density of a wave is u = \varepsilon_0 E^2 when the equal electric and magnetic contributions are combined. Averaging over a cycle replaces E^2 by its mean value E_0^2/2, so the time-averaged density becomes \langle u \rangle = \frac{1}{2}\varepsilon_0 E_0^2. Inserting \varepsilon_0 = 8.85 \times 10^{-12} and E_0 = 48\ \text{V/m} gives \langle u \rangle = \frac{1}{2}(8.85 \times 10^{-12})(48)^2 = \frac{1}{2}(8.85 \times 10^{-12})(2304) = 1.0 \times 10^{-8}\ \text{J/m}^3. The 2.0 \times 10^{-8}\ \text{J/m}^3 option forgets the factor of one-half and uses \varepsilon_0 E_0^2 directly. The 5.1 \times 10^{-9}\ \text{J/m}^3 value double-counts the averaging factor. The 4.1 \times 10^{-8}\ \text{J/m}^3 answer comes from squaring incorrectly. This averaging procedure is exactly how the NCERT and JEE syllabus convert peak fields into usable energy quantities. It is important to remember that this average already includes both the electric and magnetic contributions, since the two are equal and the compact form u = \varepsilon_0 E^2 bundles them together before averaging. Had one only counted the electric half, the result would be wrongly halved, which is the trap behind one of the distractor values. As a magnitude check, ordinary radiation has energy densities near 10^{-8}\ \text{J/m}^3, so the result is physically sensible and consistent with weak everyday fields.