(a) Since, both the blocks will move with same acceleration (say a ) in horizontal direction.
Let us take both the blocks as a system. Net external force on the system is 20 N in horizontal direction.
Using `sumF_(x)=ma_(x)`
`20=(4+2)a=6a " or "a=(10)/(3)ms^(-2)`
(b) The free body diagram of both the blocks are shown in the figure.
Using `sumF_(x)=ma_(x)`
For 4 kg block 20-N=4a=`4xx(10)/(3)`
`implies N=20-(40)/(3)=(20)/(3)N`
For 2 kg block
`N=2a=2xx(10)/(3)=(20)/(3)N`
Here, N is the normal reaction between the two blocks.
Note In free body diagram of the blocks we have not shown the forces acting on the blocks in vertical direction, because normal reaction between the blocks and acceleration of the system can be obtained without using `sumF_(y)=0`