Two blocks of $2 kg$ and $1 kg$ are in contact on a frictionless table. If a force of $3 N$ is applied on $2 kg$ block, then the force of contact between the two blocks will be

Common acceleration of the system, $a=\frac{3}{2+1}=1 m{s}^{-2}$. Let $R$ be the force of contact between the two blocks. The free body diagram of $1 kg$ block is as shown in the figure. 

The equation of motion is

$R=1\times a=1\times 1=1 N$.