CONCEPT:
Leakage current: When a capacitor is connected with a voltage V, two plates of the capacitor get charged by charges +q and -q.
No, current passes through the capacitor. Now, when a dielectric is added inside the capacitor, the charges get a medium through which
they can cross the distance between the plates and give some current as a result inside the capacitor. This current is known as the leakage current of the capacitor.

Resistivity: It is equal to the resistance of a material per unit length and per unit area.
Mathematically, Reisistance R = ρl/A
Resistivity ρ = RA/l
Capacitance: C = kϵ0A/d
where, k = dielectric constant = relative permittivity, ϵ0 = permittivity of the free space ,
d = distance between the plates, A = Area of the plates
EXPLANATION:
When the material is filled between the plates of the parallel capacitor
The capacitance C = kϵ0A/d
∴ d/A = kϵ0/C
Then the resistance of the material = R = ρd/A = ρkϵ0/C
∴ The leakage current I = V/R = V/(ρkϵ0/C) = VC/ρkϵ0
CALCULATION
Given,
V = 40 V, C = 2 pF = 2 × 10-12 F
ρ = 200 Ωm, k = 50
∴ The leakage current I = VC/ρkϵ0
⇒ I = (40×2 × 10-12) / (200×50×8.85×10-12) = 0.9 mA
Hence the correct answer is option 2.