The capacitance of a parallel plate capacitor with air as medium is $6\mu F$. With the introduction of a dielectric medium, the capacitance becomes $30\mu F$. The permittivity of the medium is :
$({\in }_0=8.85\times 10^{-12}{C}^2{N}^{-1}{m}^{-2})$ 

Capacitance of air capacitor 
$C_0=\frac{{\epsilon }_{0}A}{d}=6\mu F$           ......... (i)
When a dielectric of permittivity ${\epsilon }_r$ and dielectric constant K is introduced between the plates, then
Capacitance, $C=\frac{K{\epsilon }_{0}A}{d}=30\mu F$       ..... (ii)
Dividing eq. (ii) by (i), we get
$\frac{C}{C_0}=\frac{\frac{K{ϵ}_{o}A}{d}}{\frac{{\epsilon }_{0}A}{d}}=\frac{30}{6}$
$\Rightarrow K=5$
$\therefore$ permittivity of the medium
${\epsilon }_0={\epsilon }_{0}K$
$=8.85\times 10^{-12}\times 5=0.44\times 10^{-10}$