Two equal positive point charges are separated by a distance 2a ....

Let us assume at distance $y$, force is maximum.

Now, force at a distance $y$ is given by, ${F}_{net} = 2Fcos\theta$

So, ${F}_{net} = \dfrac{2Kq{q}_{0}y}{{\left({y}^{2} + {a}^{2}\right)}^{\dfrac{3}{2}}}$

For ${F}_{net}$ to be maximum, $\dfrac{d{F}_{net}}{dy} = 0$

Or $Kq{q}_{0}\dfrac{\left({y}^{2} + {a}^{2}\right)\dfrac{3}{2} - y\dfrac{3}{2} \times 2y\left({y}^{2} + {a}^{2}\right)\dfrac{1}{2}}{{\left({y}^{2} + {a}^{2}\right)}^{3}} = 0$

Simplifying, we get $y = \dfrac{a}{\sqrt{2}}$.

Hence, the value of $x = 2$.