Please solve the question in the image.

Assume that current I is inserted from outside into the ‘a’and withdrawn at infinity. Since the system is symmetric around the injection points the current will split equally into 2D currents in bonds connected to the junction. In particular, along the bond connecting ‘a’ and ‘b’there will be current I/(2D).
Now consider a different situation: current I is removed from ‘b’ and is injected from infinity. At each of the bonds entering ‘b’, there will be a current I/(2D).
Now consider a linear superposition of two above situations: Current I is injected into the ‘a’ and current I is withdrawn from ‘b’, and nothing happens at infinity. The current on the bond connecting ‘a’ and ‘b’ will be I/(2D)+I/(2D)=I/D. Therefore, the voltage between the junctions will be V=I/D, and consequently the effective resistance will be R=V/I=1/D.
Assume that current I is inserted from outside into the "red" junction and withdrawn at infinity. Since the system is symmetric around the injection points the current will split equally into 2D currents in bonds connected to the junction. In particular, along the bond connecting blue and red junctions there will be current I/(2D).
Now consider a different situation: current I is removed from blue junction and is injected from infinity. At each of the bonds entering the blue junction, there will be a current I/(2D).
Now consider a linear superposition of two above situations: Current I is injected into the red junction and current I is withdrawn and the blue junction, and nothing happens at infinity. The current on the bond connecting the red and blue junctions will be I/(2D)+I/(2D)=I/D. Therefore, the voltage between the junctions will be V=I/D, and consequently the effective resistance will be R=V/I=1/D. Thus, the resistance will be 1/3 for cubic, and 1/2 for square lattice.

Thanks & Regards
Bharat Bajaj
askiitian faculty
IIT Delhi