Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Recalling that the voltage across the capacitor while it is charging is given by
$$V_C=\varepsilon\left(1-e^{-t/RC}\right)\tag 1$$
Now we need to find the maximum displacement current through the capacitor where the displacement current is given by
$$I_{\rm disp }=\dfrac{dQ}{dt}$$
Recalling that $V_C=Q/C\Rightarrow Q=V_CC$,
$$I_{\rm disp }=C\dfrac{dV_C}{dt}$$
Plug from (1),
$$I_{\rm disp }=C\dfrac{d }{dt}\varepsilon\left(1-e^{-t/RC}\right)$$
$$I_{\rm disp }=\varepsilon C \left(0-\dfrac{-1}{RC}e^{-t/RC}\right)$$
$$I_{\rm disp }= \dfrac{ \varepsilon \color{red}{\bf\not} C}{R \color{red}{\bf\not} C}e^{-t/RC} $$
$$I_{\rm disp }= \dfrac{ \varepsilon }{R }e^{-t/RC}\tag 2 $$
So, the maximum current occurs when $e^{-t/RC}=1$,
Thus,
$$(I_{\rm disp })_{max}= \dfrac{ \varepsilon }{R } =\dfrac{25}{150} $$
$$(I_{\rm disp })_{max}= \color{red}{\bf 0.167}\;\rm A$$
The maximum electric flux through the capacitor is given by
$$(\Phi_e)_{max}=E_{\rm max}A$$
where $E=V_C/d$ where $d$ is the separation distance between the capacitor's plates.
$$(\Phi_e)_{max}=\dfrac{V_{\rm max}}{d}A$$
where $V_{\rm max}$ here is the voltage of the battery.
$$(\Phi_e)_{max}=\dfrac{\varepsilon }{d} A$$
Recalling that the capacitance is given by $C=\epsilon_0 A/d\Rightarrow A=Cd/\epsilon_0$
Hence,
$$(\Phi_e)_{max}=\dfrac{\varepsilon }{ \color{red}{\bf\not} d} \dfrac{C \color{red}{\bf\not} d}{\epsilon_0}$$
$$(\Phi_e)_{max}= \dfrac{\varepsilon C }{\epsilon_0}\tag 3$$
Plug the known;
$$(\Phi_e)_{max}= \dfrac{(25)(2.5\times 10^{-12}) }{(8.85\times 10^{-12})}$$
$$(\Phi_e)_{max}= \color{red}{\bf7.1}\;\rm V\cdot m$$
$$\color{blue}{\bf [b]}$$
The electric flux through the capacitor is given by
$$\Phi_e=(\Phi_e)_{\rm max}\left(1-e^{-t/RC}\right)$$
Plug from (3),
$$\Phi_e=\dfrac{\varepsilon C }{\epsilon_0}\left(1-e^{-t/RC}\right)$$
Plug the known;
$$\Phi_e=\dfrac{(25)(2.5\times 10^{-12}) }{(8.85\times 10^{-12})}\left(1-e^{-(0.5\times 10^{-9})/(150)(2.5\times 10^{-12})}\right)$$
$$\Phi_e=\color{red}{\bf 5.2}\;\rm V\cdot m$$
Now we need to plug the known into (2),
$$I_{\rm disp }= \dfrac{ 25 }{150}e^{-(0.5\times 10^{-9})/(150)(2.5\times 10^{-12})}$$
$$I_{\rm disp }=\color{red}{\bf 0.044}\;\rm A$$
