Suppose,the 44 like charges are present at A,B,C,DA,B,C,D and AB=BC=CD=DA=0.2mAB=BC=CD=DA=0.2m
We are considering forces on BB;so due to AA and CC force(FF) will be repulsive in nature along ABAB and CBCB respectively.
due to DD force (F') will also be repulsive in nature acting along diagonal DB
DB=0.2sqrt(2)m
So,F=(9*10^9*(16*10^-6)^2)/(0.2)^2 =57.6N
and F'=(9*10^9*(16*10^-6)^2)/(0.2sqrt(2))^2=28.8N
now,F' makes an angle of 45^@ with both AB and CB.
so,component of F' along two perpendicular direction i.e AB and CB will be 28.8 cos 45
So,we have two forces of (57.6+28.8 cos 45)=77.95N acting perpendicular to each other on the charge at B
So,net force on the charge at B is sqrt(77.95^2 +77.95^2)=77.95 sqrt(2)=110.24N making an angle of 45^@ w.r.t AB and CB