${C}=2 \times 10^{-12}\, {F}$

${V}=40\, {V}$

${K}=56$

$C =\frac{ K \varepsilon_0 A }{ d }$ ($K$ is the dielectric constant or relative permittivity) and

$R =\frac{\rho d }{ A }$

Now, charge will be discharged through the resistance between the plates.

Now, time constant ( $( T )$ of discharging,

$\tau= RC =\frac{\rho d }{ A } \times \frac{ K \varepsilon_0 A }{ d }$

$\tau=\rho K \varepsilon_0$

For a given R-C circuit, the discharged current is given by

$i =\frac{ Q }{ RC } e ^{-\frac{ t }{ RC }}$

$i =\frac{ Q }{ pK \varepsilon_0} e ^{-\frac{ t }{ pK \varepsilon_0}}$

The above discharge current is the leakage current,

$i _{\text {leakage }}=\frac{ Q }{\rho K \varepsilon_0} e ^{-\frac{ t }{\rho K \varepsilon_0}}$

Maximum leakage current,

$\left( i _0\right)_{\text {leakage }} =\frac{ Q }{\rho K \varepsilon_0}=\frac{ CV }{\rho K \varepsilon_0}$

$=\frac{2 \times 10^{-12} \times 40}{200 \times 50 \times 8.85 \times 10^{-12}}$

$=903 \mu A =0.9 mA$

$\left( i _0\right)_{\text {leakage }} =0.9 \;mA$