F = (M+M)a                 

 a=F/m+M ............1

bw m and M there is normal reaction let it be N,

 then  at m

                 F-N=ma

                N=F-ma  or

                  =F(1 - m/(m+M) )

                  N=F( M/m+M)................2

 if both blocks are at rest wrt each other then it can be said that friction is sufficient to balance the weight of  block of mass m..

   (friction force) f =mg

             f =u(N)                    (u is cofficient of friction and N is normal reaction)

              uN=mg

              u(F)(M/m+M) =mg

                 F=mg(m+M)/uM

  on solving

                    F=240N

this is the minimum force required for which block remmains in equilibrium...

but the answer was given as 10 n i got the same answer