A parallel plate capacitor of capacity $100pF$ is connected to $230V$ of AC supply of $300rad/s$ frequency. The value of displacement current is

Displacement current, 
${\mathrm{I}}_{\mathrm{D}}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{\mathrm{V}}{{\mathrm{X}}_{\mathrm{C}}}$  where Z is impedance and $X_C$ is capacitive reactance.

${\mathrm{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega c}}$
$\begin{array}{rcl} & \Rightarrow & {\mathrm{I}}_{\mathrm{D}}=\frac{\mathrm{V}}{1/\mathrm{\omega c}}=\mathrm{V}\times \mathrm{\omega c}\\ & \Rightarrow & {\mathrm{I}}_{\mathrm{D}}=230\times 300\times 100\times {10}^{-12}\\ & \Rightarrow & {\mathrm{I}}_{\mathrm{D}}=69\times {10}^{-7}=6.9 \mathrm{\mu A}\end{array}$