See in this case, accelaration = 3m/s2,
Frictional force =

(m1 + m2 +m3 ) g .
The external force F has to overcome this frictional force and additionally provide the extra force to give accelaration of 3m/s2.
Total mass = 4.5kg.
Total foce for accelaration = m x a = 4.5 x 3 = 13.5N
The total frictional force : 4.5 x 9.8 x 0.2 = 8.82
Therefore total force applied F = 22.5 N .
Now the equation for m1 is = F – R = m1a
Where R is the reaction force from m2 on m1 .
22.5 – R = 1.5 x 3 so R = 18N and since R12 =R21 therefore the force on m2 due to m1 should also be 18.
Since that is not an option, you should go with 15N as that is the closest to this value.
This is what i think it should be but i am a little confused. From another method the answer is coming 9N so tell me what the correct answer is.