b
\(\begin{array}{l}
Here,\,{M_A} = 4\,kg,\,{M_B} = 2kg,\,{M_c} = 1\,kg,\,F = 14\,N\\
Net\,mass,\,M = {M_A} + {M_B} + {M_C} = 4 + 2 + 1 = 7kg\\
Let\,a\,be\,the\,acceleration\,of\,the\,system.\\
{\rm{Using}}\,Newton's\,{\rm{second}}\,law\,of\,motion,\\
\,\,\,\,\,\,\,\,\,\,\,F = Ma\\
\,\,\,\,\,\,\,\,\,\,\,14\, = 7a\,\,\,\,\therefore \,\,\,\,a = 2\,\,m{s^{ - 2}}
\end{array}\)

\(\begin{array}{l}
Let\,\,F\,be\,the\,force\,applied\,on\,block\,A\,by\,block\,B\\
i.e,.\,the\,constact\,force\,between\,A\,and\,B.\,Free\,body\,\\
diagram\,for\,block\,A\\
Again\,u{\rm{sing Netwton's}}\,{\rm{second}}\,{\rm{law}}\,{\rm{of}}\,{\rm{motion,}}\\
\,\,\,\,\,\,\,\,\,\,{\rm{F - F' = 4a}}\\
\,\,\,\,\,\,14 - F'\, = 4 \times 2\,\, \Rightarrow 14 - 8 = F\,\,\,\therefore F = 6N
\end{array}\)