Let a be the side of the square plate.  
As shown in figure, $C_1$ and $C_2$ are in parallel. Therefore,  total capacity of capacitors in the position shown is 
$C = C_1 + C_{2 }$
 C =  $\frac{{\in }_{0}a(a-x)}{d}+\frac{K{\in }_{0}ax}{d}$
$\therefore$ q = CV = $\frac{{\in }_{0}aV}{d}$  (a – x + Kx)
As plates are lowered in the oil, C increases hence charge stored will increase. 
Therefore, i =$\frac{dq}{dt}$  =$\frac{{\in }_{0}aV}{d}$   (K–1) ·  $\frac{dx}{dt}$
Substituting the values
 ${\in }_0= 8 \times {10}^{–12} C^2/N–m^2$
a = 1m, V = 500 volt, d = 0.01m, K = 11 and $\frac{dx}{dt}$  = speed of plate = 0.001 m/s  
We get current  i =  $\frac{(8\times {10}^{-12}) 1 \left(500\right) (11-1 ) \left(0.001\right)}{\left(0.01\right)}$  Amp.
 i = 4