Placing a charge

Let us place the third charge at a point $(x,0)$ such that $0 < x <3$ since otherwise the field direction would not oppose one another. (*)

So now the net force on this third charge would in fact be the sum of the two electrostatic forces. To be specific, remark that we will be using $x$ to calculate the force between the first charge and this third charge and we will be using $3 - x$ to calculate the force between the first charge and the second charge.

Now, since we are given the net force on the third charge is 0, these two forces stated in the above paragraph need to be equal. We will apply Coulomb's Law:

$\frac{k \times 4Q \times Q_2}{x^{2}}$ = $\frac{k \times Q \times Q_2}{(3-x)^{2}}$

After simplifying we get $\frac {4Q}{x^{2}} = \frac {Q}{(3-x)^{2}}$ .

Further rearranging will give us $x^2-8x+12=0$ which factors as $(x-2)(x-6)=0$ This gives rise to the solution of $x = 2$ or $6$ However due to statement (*), we can reject $6$ and hence the answer is 2