The energy associated with electric field is $$(U_E)$$ and with magnetic field is $$(U_B)$$ for an electromagnetic wave in free space. Then:

The energy stored per unit volume (energy density) in an electric field of magnitude $$E$$ in free space is given by the well-known electrostatic formula

$$u_E \;=\;\dfrac{1}{2}\,\varepsilon_0\,E^{\,2},$$

where $$\varepsilon_0$$ is the permittivity of free space.

In exactly the same way, the energy density in a magnetic field of magnitude $$B$$ is obtained from magnetostatics as

$$u_B \;=\;\dfrac{1}{2\mu_0}\,B^{\,2},$$

with $$\mu_0$$ being the permeability of free space.

For a plane electromagnetic wave travelling in free space, the magnitudes of its electric and magnetic fields are not independent; they obey the intrinsic relation

$$E \;=\;c\,B,$$

where $$c$$ is the speed of light in vacuum. We shall now exploit this relation to compare $$u_E$$ and $$u_B$$.

First, solve the above relation for $$B$$:

$$B \;=\;\dfrac{E}{c}.$$

We substitute this expression for $$B$$ into $$u_B$$:

$$u_B \;=\;\dfrac{1}{2\mu_0}\,\Bigl(\dfrac{E}{c}\Bigr)^{\!2} \;=\;\dfrac{1}{2\mu_0}\,\dfrac{E^{\,2}}{c^{\,2}}.$$

Now, recall the defining equation that links the electromagnetic constants with the speed of light,

$$c^{\,2} \;=\;\dfrac{1}{\mu_0\,\varepsilon_0}.$$

Therefore, its reciprocal is

$$\dfrac{1}{c^{\,2}} \;=\;\mu_0\,\varepsilon_0.$$

Substituting $$1/c^{\,2} = \mu_0\varepsilon_0$$ into the expression for $$u_B$$ gives

$$u_B \;=\;\dfrac{1}{2\mu_0}\,E^{\,2}\,(\mu_0\varepsilon_0) \;=\;\dfrac{1}{2}\,\varepsilon_0\,E^{\,2}.$$

But this is precisely the expression we obtained earlier for $$u_E$$. Hence,

$$u_E \;=\;u_B.$$

The energy carried by an electromagnetic wave in free space is therefore shared equally between its electric and magnetic fields.

Hence, the correct answer is Option B.