1. For an electromagnetic wave in free space, the instantaneous energy densities are:

$u_E = \frac{1}{2}\varepsilon_0 E^2$

and

$u_B = \frac{B^2}{2\mu_0}$

2. Given:

$E = E_0 \sin(\omega t-kx), \qquad B = B_0 \sin(\omega t-kx)$

So the average values over one complete cycle are:

$\langle u_E \rangle = \frac{1}{2}\varepsilon_0 \langle E^2 \rangle = \frac{1}{2}\varepsilon_0 E_0^2 \langle \sin^2(\omega t-kx) \rangle$

$\langle u_B \rangle = \frac{1}{2\mu_0} \langle B^2 \rangle = \frac{1}{2\mu_0} B_0^2 \langle \sin^2(\omega t-kx) \rangle$

3. Since the time average of $\sin^2$ over a full cycle is:

$\langle \sin^2 \theta \rangle = \frac{1}{2}$

we get:

$\langle u_E \rangle = \frac{1}{4}\varepsilon_0 E_0^2$

$\langle u_B \rangle = \frac{1}{4\mu_0} B_0^2$

4. Therefore,

$\frac{\langle u_E \rangle}{\langle u_B \rangle} = \frac{\frac{1}{4}\varepsilon_0 E_0^2}{\frac{1}{4\mu_0} B_0^2} = \mu_0\varepsilon_0 \frac{E_0^2}{B_0^2}$

5. For a plane electromagnetic wave in free space,

$E_0 = cB_0$

and

$c^2 = \frac{1}{\mu_0\varepsilon_0}$

Hence,

$\mu_0\varepsilon_0 \frac{E_0^2}{B_0^2} = \mu_0\varepsilon_0 c^2 = 1$

6. So the ratio of average electric energy density to average magnetic energy density is:

$\boxed{1}$

Option A is correct.