= Solution
By reflection symmetry the vertical <gravitational field> of the <self-gravitating incompressible slab> vanishes at its mid-plane. Inside the slab, the gravitational <Poisson equation> therefore integrates to $d\psi/dz=4\pi G\rho z$. Treat the <star> as a test particle, so its own <mass> does not alter that field. Its <vertical stellar orbit in a uniform slab> obeys
$$
\ddot z=-4\pi G\rho z,\qquad z(0)=a,\qquad\dot z(0)=0.
$$
Hence $z(t)=a\cos(\omega t)$ with $\omega=\sqrt{4\pi G\rho}$. The first mid-plane crossing occurs after one quarter of this <simple harmonic motion> period:
$$
\boxed{t_{\rm cross}=\frac{\pi}{2\sqrt{4\pi G\rho}}=\frac{\sqrt\pi}{4\sqrt{G\rho}}.}
$$
Its independence of $a$ follows from the linear interior <restoring force>. The calculation neglects drag and changes to the slab, and it uses the infinite planar approximation rather than the field of a finite Galactic disk.
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