Answer
$\frac{Q}{q} = -2\sqrt 2$
Work Step by Step
$F_{on. 1} = F_{2 on 1} + F_{3on1} + F_{4on1}$
Since $q_1 = q_2$ the charges have the same sign and repel each other
The formula for the electrostatic force of $q_4$ on $q_1$:
$ F_{4on1} = \frac{Kq_1q_2}{r^2_{14}}$
${r^2_{14}} = a^2 +a^2 = 2a^2$
$ F_{4on1,x} = \frac{Kq_1q_2}{r^2_{14}}(-cos(45))$
$ F_{4on1,y} = \frac{Kq_1q_2}{r^2_{14}}(sin(45))$
$ F_{4on1} = \frac{Kq_1q_2}{r^2_{14}}(-cos(45) +sin(45))$
$ F_{4on1} = \frac{KQ^2}{2a^2}(-\frac{1}{\sqrt 2} i+\frac{1}{\sqrt 2} j)$
$ F_{4on1} = \frac{KQ^2}{2\sqrt 2 a^2}(- i+j)$
In order for $F_{2on1}$ must be in the +$i$ direction to cancel the x component on $F_{4on1}$ and $F_{3on1}$ must be in the -$ j $
This means that charges Q and q must have opposite signs
$ F_{2on1} = \frac{Kq_1q_2}{r^2_{12}}$
$r^2_{12} = a^2$
$ F_{2on1} = \frac{Kq_1q_2}{a^2}$
$ F_{2on1} = \frac{kQq}{a^2}(-i)$
$ F_{3on1} = \frac{KQq}{a^2}(-j)$
For the net to be zero the x components must be added together to equal zero:
$ F_{2on1} + F_{4on1,x} = \frac{KQq}{a^2}(-i) + \frac{KQ^2}{r^2_{14}}(-cos(45)) = 0$
$\frac{KQ}{a^2}(q-\frac{Q}{2\sqrt 2 }) = 0$
$q-\frac{Q}{2\sqrt 2 } = 0$
$\frac{|Q|}{|q|}= 2\sqrt 2$
Since Q and q have opposite signs the answer must be negative:
$\frac{|Q|}{|q|}= -2\sqrt 2$
