When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are, respectively, 0.6 V and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then

Stopping  potential depends on frequency and not on the intensity and the source is same and stopping potentials remains unaffected.
 $I\propto \frac{1}{r^2}$
$Saturation current\propto I$
 $\therefore \frac{{\text{(Saturation current)}}_{\mathrm{final}}}{{\text{(Saturation current)}}_{\mathrm{Inital}}}= \frac{{r^2}_{\left(initial\right)}}{{r^2}_{\left(final\right)}}$  
(  Saturation current )_final =  $\frac{\text{0.2×0.2}}{\text{0.6×0.6}}\text{×18}$  = 2 mA