Capacitance of parallel plate capacitor with air between the plates $C=\frac{A{\epsilon }_0}{d}$; where $A$ is plate area and $d$ is separation between the plates.

As we know that $Q = CV$

Initial charge 

$Q_i = 90\times {10}^{-12}\times 20=1800\times {10}^{-12} \mathrm{C}$

$\Rightarrow Q_i=1.8 \mathrm{nC}$

Dielectric material of dielectric constant $k=\frac{5}{3}$ is inserted between the plates, then capacitance becomes

$C\text{'}=\frac{A{\epsilon }_{0}k}{d}$

$\Rightarrow C\text{'}=kC=\frac{5}{3}C$

$\Rightarrow C\text{'}=\frac{5}{3}\times 90\times {10}^{-12}=150\times {10}^{-12} \mathrm{F}$

Now the final charge $Q_f=C\text{'}V$

$\Rightarrow Q_f=150\times {10}^{-12}\times 20=3000\times {10}^{-12} \mathrm{C}$

$\Rightarrow Q_f=3 \mathrm{nC}$

Induced charge,

 $Q_{\mathrm{induced}}=Q_f-Q_i$

$Q_{\mathrm{induced}}=3-1.8=1.2 \mathrm{nC}$.