= Solution
Let $\rho_0$ denote the midplane <mass density> and choose the symmetry plane at $z=0$. Isothermal <hydrostatic equilibrium> and the <Poisson equation for Newtonian gravity> give
$$
c_s^2\frac{d\ln\rho}{dz}=-\frac{d\Phi_0}{dz},\qquad
\frac{d^2\Phi_0}{dz^2}=4\pi G\rho,
$$
so $c_s^2(d^2/dz^2)\ln\rho=-4\pi G\rho$. For $\rho=\rho_0\operatorname{sech}^2(z/H)$, direct differentiation yields
$$
\frac{d\ln\rho}{dz}=-\frac2H\tanh(z/H),\qquad
\frac{d^2\ln\rho}{dz^2}=-\frac2{H^2}\operatorname{sech}^2(z/H).
$$
Thus the profile is a <self-gravitating isothermal slab> when $H^2=c_s^2/(2\pi G\rho_0)$. Integrating over both sides gives $\Sigma=2\rho_0H$, hence
$$
\boxed{H=\frac{c_s^2}{\pi G\Sigma},\qquad
\rho_0=\frac{\pi G\Sigma^2}{2c_s^2}}.
$$
The corresponding potential is $\Phi_0(z)-\Phi_0(0)=2c_s^2\ln\cosh(z/H)$, whose <gradient> vanishes at the midplane and tends to $\pm2\pi G\Sigma$ far from the slab. The profile parameter $H$ is not its asymptotic exponential <mass density> <scale height>: that is $H/2$.
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