In an electromagnetic wave propagating in vacuum, what fraction of...

In an electromagnetic wave, the energy is equally divided between the electric and magnetic fields.

The average energy density of the electric field ($u_E$) is equal to the average energy density of the magnetic field ($u_B$).

Total average energy density, $u = u_E + u_B = 2u_E$

Therefore, the fraction of the total energy density contributed by the electric field is $\dfrac{u_E}{u} = \dfrac{u_E}{2u_E} = \dfrac{1}{2}$.

Answer: $\dfrac{1}{2}$