Answer
See the detailed answer below.
Work Step by Step
We know that the displacement current is given by
$$I_{\rm displacement }=\epsilon_0\dfrac{d\Phi_e}{dt}\tag 1$$
Recalling that the electric flux inside a capacitor is given by
$$\Phi_e=EA\tag 2$$
where $A$ is the plate's area, and $E$ is the electric field between the two plates which is given by
$$E=\dfrac{Q}{A\epsilon_0}$$
where $Q= V_CC$,
$$E=\dfrac{ V_CC}{A\epsilon_0}$$
Plug into (2),
$$\Phi_e=\dfrac{ V_CC}{A\epsilon_0}A =\dfrac{ V_CC}{ \epsilon_0}$$
Plug into (1),
$$I_{\rm displacement }=\epsilon_0\dfrac{d}{dt}\dfrac{ V_CC}{ \epsilon_0}$$
$$\boxed{I_{\rm displacement }=C \dfrac{dV_C}{dt} }$$
