As shown in the figure, after passing through the medium 1, the speed of light $$v_2$$ in medium 2 will be: (Given $$c = 3 \times 10^8$$ m s$$^{-1}$$)

We need to find the speed of light $$v_2$$ in medium 2, given the properties shown in the diagram.

The refractive index $$\mu$$ of a medium is related to its relative permittivity $$\varepsilon_r$$ and relative permeability $$\mu_r$$ by the relation: $$\mu = \sqrt{\mu_r \varepsilon_r}$$.

From the given figure, both medium 1 and medium 2 have a relative permeability of $$\mu_r = 1$$.

For medium 2, the relative permittivity is given as $$\varepsilon_r = 9$$. Therefore, the refractive index of medium 2 is: $$\mu_2 = \sqrt{1 \times 9} = 3$$.

The refractive index of a medium is also defined as the ratio of the speed of light in vacuum $$c$$ to the speed of light in that medium $$v$$. For medium 2, this is given by: $$\mu_2 = \frac{c}{v_2}$$.

Rearranging the formula to solve for $$v_2$$ yields: $$v_2 = \frac{c}{\mu_2}$$.

Substituting the given values $$c = 3 \times 10^8\text{ m s}^{-1}$$ and $$\mu_2 = 3$$ into the expression gives: $$v_2 = \frac{3 \times 10^8}{3} = 1.0 \times 10^8\text{ m s}^{-1}$$.

Therefore, the correct answer is Option A: 1.0 × 108 m s-1.