The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability for this wavelength, will be:

The correct option is 30°.

To find the critical angle of the medium, we first determine its refractive index (represented by $n$). The refractive index of a electromagnetic medium is related to its relative permittivity (${\epsilon }_r$) and relative permeability (${\mu }_r$) by the following relation:

$$n=\sqrt{{\epsilon }_r{\mu }_r}$$

Given that the relative permittivity of the medium is ${\epsilon }_r=3$ and the relative permeability for this wavelength is ${\mu }_r=\frac{4}{3}$, we can substitute these values into the formula:

$$n=\sqrt{3\times \frac{4}{3}}=\sqrt{4}=2$$

The relationship between the critical angle (${\theta }_c$) of a medium and its refractive index ($n$) when light travels from the medium into air or vacuum is given by:

$$sin\left({\theta }_c\right)=\frac{1}{n}$$

Substituting the calculated value of $n=2$ into this relation:

$$sin\left({\theta }_c\right)=\frac{1}{2}$$

Taking the inverse sine of both sides, we find the critical angle:

$${\theta }_c={sin}^{-1}\left(\frac{1}{2}\right)=30°$$

Therefore, the critical angle of the medium for this wavelength is 30°.