Three blocks A,B and C of masses 5kg, 3kg and 2kg respectively are kept in contact on a smooth horizontal surface. Two horizontal forces 20N and 4N act on A and C respectively in opposite directions. The contact force between B and C is

Since, the blocks are in contact on a smooth horizontal surface, they will a single system. For this system, acceleration will be

$\mathrm{a}=\left(\frac{20-4}{5+3+2}\right){\mathrm{ms}}^{-2}=1.6{\mathrm{ms}}^{-2}$

For the block C, on one side there will be the contact force (force between B and C) and on the other side there will be the applied force of 4N. 

According to Newton's second law, 

${\mathrm{F}}_{\mathrm{net}}=\mathrm{ma}$

$\text{⇒ }2\mathrm{a}={\mathrm{f}}_{\mathrm{c}}-4$

$\Rightarrow {\mathrm{f}}_{\mathrm{c}}=(2\times 1.6+4)\mathrm{N}=7.2\mathrm{N}$