I1=30 AI=10 AI2=20 A 
Also given; length of wire Q =25 cm=0.25 m Force on wire Q due to wire R FQR=10−7×0.052×20×10×0.25=20×10−5 N (Towards left) Force on wire Q due to wire P FQP=10−7×0.032×30×10×0.25=50×10−5 N (Towards right) Hence, Fnet =FQP−FQR=50×10−5 N−20×10−5 N=3×10−4 N towards right