In order to establish an instantaneous displacement current of 1 mA in the space between the plates of 2µF parallel plate capacitor, the potential difference need to apply is:

500 Vs^-1

$I_{D}=1 mA=10^{-3} A, C=12 \mu F=2 × 10^{-6} F$

$I_{D}=I_{C}=\frac{d}{dt}(CV)=C \frac{dV}{dt}$

Therefore, $\frac{dV}{dt}=\frac{I_{D}}{C}=\frac{10^{-3}}{2 × 10^{-6}}=500 Vs^{-1}$

Therefore, applying a varying potential difference of 500 Vs^-1 would produce a displacement current of desired value.