Since AB=BC=DB=L, we have AD = √2L
Now force between particles placed at A and D is: F=Gmm/(√2L)2=Gm2/2L2
Similar force particle at C will exert on particle at D.
So total force on particle D will be 2F.
Now the resultant component of these two forces along DB will be: 2Fcos 45∘
Since AB=BC=BD=L and ∠DBC=∠DBA=90∘, we get ∠ADB=∠BDC=45∘
So, the resultant force on mass m at D will be Gm2√2L2