Consider the diagram below. We can assume that Q1 = 6nC, Q2 = -9nC and Q3 = 5nC since the question is incomplete
F31=r2kq1q3=(2×10−4)29×109×(6×10−9)(5×10−9)=6.75N
F32=r2kq2q3=(4×10−4)29×109×(−9×10−9)(5×10−9)=−2.53125N
Net force on Q3 =F32+F31=6.75+−2.53125=4.21875N
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