Force on charge q1 due to charge q2 is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
[tex]F = \frac{9*10^9 * 3*10^{-6}*5 * 10^{-6}}{0.2^2}[/tex]
[tex]F = 3.375 N[/tex]
Now the net force on q1 is 7 N so the force due to q3 must be
[tex]F = 7 - 3.375 = 3.625 N[/tex]
[tex]F = \frac{kq_1q_3}{r^2}[/tex]
[tex]3.625 = \frac{9*10^9 * 3*10^{-6}*8*10^{-6}}{r^2}[/tex]
by solving above
[tex]r = 0.24 m[/tex]
so q3 is at distance 0.24 m from q1