B

To find the index of refraction of a medium in terms of the electric permittivity (ε) and magnetic permeability (μ), we can use the relationship defined in optics. The index of refraction n is given by:

$$ n = rac{c}{v} $$
where:
- $c$ is the speed of light in a vacuum, and
- $v$ is the speed of light in the medium.

The speed of light in a vacuum is given by:

$$ c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} $$
and the speed of light in the medium is given by:

$$ v = \frac{1}{\sqrt{\epsilon \mu}} $$

Substituting these into the equation for n gives us:

$$ n = \frac{c}{v} = \frac{\frac{1}{\sqrt{\epsilon_0 \mu_0}}}{\frac{1}{\sqrt{\epsilon \mu}}} = \sqrt{\frac{\epsilon \mu}{\epsilon_0 \mu_0}} $$
Therefore, the correct expression for the index of refraction of the medium in terms of the permittivity and permeability of free space is:

$$ n = \sqrt{\frac{\epsilon}{\epsilon_0} \cdot \frac{\mu}{\mu_0}} $$
Hence, the answer is option B.