Hey there!:
q1 = 4.25 * 10^-6 C is at x1 = 0 m
q2 = - 5.75 * 10^-6 C is at x2 = 0.23 m
q3 = - 9 uC = -9 * 10^-6 C
for the net force on q1 to be in -x direction
q3 must be placed to the left of q1
the net force on q1 , F = k * q1 * q3 /( x3^2) - k * q1 * q2 /( x2 - x3)^2
6 = 9 * 10^9 * 10^-6 * 4.25 * 10^-6 * (9 /x3^2 - 5.75 /( 0.25 - x3 )^2 )
solving for x3
x3 = - 0.22 m
the charge q3 is placed at x3 = - 0.22 m
Hope this helps!