= Solution
For a displacement $(0,y)$ of the middle wire, the horizontal attractions towards the two fixed outer wires cancel. Each attraction has magnitude $\mu_0I^2/(2\pi\sqrt{d^2+y^2})$ per unit length. Projecting them in the $y$ direction gives
$$
\mu_m\ddot y=-\frac{\mu_0I^2}{\pi}\frac{y}{d^2+y^2}
\simeq-\frac{\mu_0I^2}{\pi d^2}y\qquad(|y|\ll d).
$$
Neglecting dissipative effects and other transverse <forces>, this is <simple harmonic motion>. Release from rest at $y_0$ gives
$$
\boxed{y(t)=y_0\cos\Omega t,\qquad
\Omega=\sqrt{\frac{\mu_0I^2}{\pi\mu_m d^2}}=2.00\,\mathrm{s^{-1}},\qquad
T=\frac{2\pi}{\Omega}=3.14\,\mathrm s.}
$$
The inverse-angular-frequency timescale is $\Omega^{-1}=0.500\,\mathrm s$; it differs from the oscillation period by $2\pi$. The assumed displacement plane matters: small displacements towards one outer wire along $x$ are unstable, rather than oscillatory.
This result concerns the specified magnetic motion with gravity absent from, or compensated in, the $y$ direction. The source diagram fixes coordinate directions, not their orientation relative to gravity. If $y$ is physically vertical with upward positive and the wire is unsupported, add $-\mu_m g$ to the <force> per unit length. The maximum restoring magnetic <force> is only $\mu_0I^2/(2\pi d)=2.0\times10^{-5}\,\mathrm{N\,m^{-1}}$, compared with weight $9.8\times10^{-3}\,\mathrm{N\,m^{-1}}$; that different setup would fall and could not oscillate about the original plane.
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