Answer
(a) $Q_o = 0.36 \mathrm{~n C}$
(b) $Q_K = 0.97 \mathrm{~n C}$
Work Step by Step
(a) the capacitance between the two plates occur due to the accumulation of the charges on the plates because of the potential and for the air, it is given by
$$Q_o = C_oV$$
Where $V_{ab} $ is the potential between the two plates and is calculated by
$$V = Ed = ( 3 \times 10^{4} \,text{V/m})(0.0015 \,\text{m}) = 45 \,\text{V}$$
Let us substitute the values $C$ and $V$ to get the charge on each plate
$$Q_o = C_oV = (8\times 10^{-12} \,\text{F})(45 \,\text{V}) = \boxed{0.36 \mathrm{~n C}}$$
(b) After the dielectric material with a dielectric constant is added $K = 2.70 $ the charges will be given by
$$Q_K = K C_oV = 2.70 (8\times 10^{-12} \,\text{F})(45 \,\text{V}) = \boxed{0.97 \mathrm{~n C}}$$
