Answer
(a) $\Delta Q = 6.3\times 10^{-6} \,\text{C}$
(b) $Q_i = 6.3\times 10^{-6} \,\text{C}$
(c) Myler sheet has no effect on the electric field
Work Step by Step
The charge $Q$ after inserting the Myler is given by
$$Q = KQ_o$$
Where $$Q_o = C_o V =(0.25 \times 10^{-6} \,\text{F}) (12 \,\text{V}) = 3 \times 10^{-6} \,\text{C}$$
Therefore we get $Q$ by
$$Q = KQ_o = 3.1 ( 3 \times 10^{-6} \,\text{C}) = 9.3 \times 10^{-6} \,\text{C}$$
The additional charge $\Delta Q$ will be
$$\Delta Q = Q - Q_o = 9.3 \times 10^{-6} \,\text{C} - 3 \times 10^{-6} \,\text{C} = \boxed{6.3\times 10^{-6} \,\text{C}}$$
(b) The total induced charges $Q_i$ on Myler surface is the same additional charges after inserting Myler sheet and it would be
$$\boxed{Q_i = 6.3\times 10^{-6} \,\text{C}}$$
(c) The electric field will not change because the induced charge is the same for the additional charges. Hence, Myler sheet has no effect on the electric field
