The displacement current of 4 . 425   μA is developed in the space between the plates of parallel…

Displacement current is given by $i_d=$$\frac{{\epsilon }_{0}d{\varphi }_E}{dt}$

Or $i_d=\frac{{\epsilon }_{0}d}{dt}\left(EA\right)$, where $E=\frac{q}{A{\epsilon }_0}$

Or $i_d={\epsilon }_0\frac{d}{dt}\left(\frac{qA}{A{\epsilon }_0}\right)=\frac{dq}{dt}=\frac{d}{dt}\left(CV\right)$

Or $i_d=C\frac{dV}{dt}=\frac{{\epsilon }_{0}A}{d}\frac{dV}{dt}$

Putting the values, we have 

$4.425\times {10}^{-6}=\frac{8.85\times {10}^{-12}\times 40\times {10}^{-4}\times {10}^6}{ d}$

$d=2\times {10}^{-6}\times {10}^{-4}\times {10}^6\times 40$

$d=80\times {10}^{-4}=8\times {10}^{-3} \mathrm{m}$

Hence, value of $x=8$.