To solve this problem, we need to understand the relationship between the capacitance of a capacitor,$\\$the dielectric constant, and the permittivity of the medium.$\\$The capacitance of a parallel plate capacitor without a dielectric is given by:$\\ C_0 = \frac{\epsilon_0 A}{d} \\$where $C_0$ is the initial capacitance, $\epsilon_0$ is the permittivity of free space,$\\ A$ is the area of the plates, and $d$ is the separation between the plates.$\\$When a dielectric medium is introduced, the capacitance becomes:$\\ C = \frac{\epsilon A}{d} \\$where $\epsilon$ is the permittivity of the dielectric medium.$\\$The relationship between the capacitance with and without the dielectric is:$\\ C = K \cdot C_0 \\$where $K$ is the dielectric constant of the medium.$\\$Given:$\\ C_0 = 6 \mu F = 6 \times 10^{-6} F \\ C = 30 \mu F = 30 \times 10^{-6} F \\$We can find the dielectric constant $K$ as follows:$\\ K = \frac{C}{C_0} = \frac{30 \times 10^{-6}}{6 \times 10^{-6}} = 5 \\$Now, we know that $\epsilon = K \cdot \epsilon_0. \\$Substitute the values:$\\ \epsilon = 5 \cdot 8.85 \times 10^{-12}$ C$^2$ N$^{-1}$ m$^{-2} \\ \epsilon = 44.25 \times 10^{-12}$ C$^2$ N$^{-1}$ m$^{-2} \\$Convert this to the correct format:$\\ \epsilon = 0.44 \times 10^{-10}$ C$^2$ N$^{-1}$ m$^{-2} \\$Therefore, the permittivity of the medium is $0.44 \times 10^{-10}$ C$^2$ N$^{-1}$ m$^{-2}. \\$This corresponds to Option 2.