d
Charge on Capacitor, $\mathrm{Q}_{i}=\mathrm{CV}$
After inserting dielectric of dielectric constant = $\mathrm{K}$
$\mathrm{Q}_{\mathrm{f}}=(\mathrm{k} \mathrm{C}) \mathrm{V}$
Induced charges on dielectric
$ \mathrm{Q}_{\mathrm{ind}} =\mathrm{Q}_{\mathrm{f}}-\mathrm{Q}_{\mathrm{i}}=\mathrm{KCV}-\mathrm{CV} $
$=(\mathrm{K}-1) \mathrm{CV}=\left(\frac{5}{3}-1\right) \times 90 \,\mathrm{pF} \times 2\, \mathrm{V}=1.2\, \mathrm{nc} $