Answer
a) $\approx 83\;\rm \mu C/m^2$
b) $\approx 14\;\rm \mu C/m^2$
Work Step by Step
We have here a parallel plate capacitor consisting of two disks and a Pyrex glass as a dielectric material between them.
$$\color{blue}{\bf [a]}$$
We know that the charge density is given by
$$\eta=\dfrac{Q}{A}$$
where $Q$ is the charge on the plates and it is given by $Q=\pm C\Delta V_C$
$$\eta=\dfrac{\pm C\Delta V_C}{A}$$
where $C=\kappa C_0$ since we have a dielectric material.
$$\eta=\dfrac{\pm \kappa C_0\Delta V_C}{A}$$
where $C_0=\dfrac{\epsilon_0 A}{d}$
$$\eta=\dfrac{\pm \kappa \epsilon_0 \color{red}{\bf\not} A\Delta V_C}{ \color{red}{\bf\not} Ad}$$
$$\eta=\dfrac{\pm \kappa \epsilon_0 \Delta V_C}{ d}$$
Plug the known;
$$\eta=\dfrac{\pm (4.7)(8.85\times 10^{-12})(1000) }{(0.5\times 10^{-3})}$$
$$\eta=\pm \color{red}{\bf 83.2}\;\rm \mu C/m^2$$
$$\color{blue}{\bf [a]}$$
We know that the induced surface charge density on the dielectric material between the two plates capacitor is given by
$$\eta_{\rm induced}=\eta_0\left[ 1-\dfrac{1}{\kappa} \right]$$
where $\eta_0=\eta/\kappa$, so
$$\eta_{\rm induced}=\dfrac{\eta}{\kappa}\left[ 1-\dfrac{1}{\kappa} \right]$$
$$\eta_{\rm induced}=\dfrac{\eta}{\kappa}\left[ \dfrac{\kappa-1}{\kappa} \right]$$
$$\eta_{\rm induced}= \eta\left[ \dfrac{\kappa-1}{\kappa^2} \right]$$
Plug the known;
$$\eta_{\rm induced}= (83.2\times 10^{-6})\left[ \dfrac{(4.7)-1}{(4.7)^2} \right]$$
$$\eta_{\rm induced}=\pm \color{red}{\bf 13.9}\;\rm \mu C/m^2$$
