Your calculation of $T$ is correct - and possibly simpler than integration, which is the usual method for finding the electrostatic repulsion force $F$. Perhaps your difficulty is that the result which you get for the amplitude does not match the answer which you have been given for $x$?
The wire is initially in equilibrium between two forces : the repulsive force $F$ between the two charges $+Q$ and $+q$ which pushes the wire to the right, and the 2 tensions $T$ in each spring which pull the wire to the left : $F=2T$. The initial extension of the springs is the amplitude of oscillations : $T=kA$.
You are not asked to find the period of oscillations, so it is not necessary to write the equation of motion and deduce the angular frequency. To find the period we would need to know the mass of the wire, which is not stated. The amplitude will be the same regardless of the mass of the wire.
You have found $T$. Finding the amplitude $A$ is then straightforward. If the result $x=2a$ is not what you expected, there may be an error in the question or in the given answer.
$qQ/4\pi \epsilon_0 a^2$ has the unit of force ($N$), and $k$ has unit of force per length ($N/m$). So in order for $xqQ/8\pi^2\epsilon_0 a^3 k$ to have the unit of length (amplitude) then $x$ must also have the unit of length.
Note : Do not confuse $x$ with the extension of the springs. In this question $x$ is an unknown factor in an equation. The initial extension of the springs is the amplitude $A$.