Given,
The charges are:
[tex]q_1=5.0\mu C\text{ at x=20 }cm[/tex][tex]q_2=4.0\mu C\text{ at x=0 }cm[/tex][tex]q_3=10\mu C\text{ at x=-30 }cm[/tex]The distancce between q1 and q2 is 20 cm
Thus,
The force on q2 due to q1 is:
[tex]F=k\frac{5\times4}{20^2}=8.9\times10^9\times\frac{20}{20^2}=0.00045C[/tex]