= Solution
Neglect disk self-<gravity> and take $|z|\ll r$ around a central <point mass> $M_\star$. Its vertical gravitational <acceleration> is
$$
g_z=-\frac{GM_\star z}{(r^2+z^2)^{3/2}}\simeq-\frac{GM_\star}{r^3}z=-\Omega^2z.
$$
Vertical <hydrostatic equilibrium> is $dP/dz=\rho g_z$. For a vertically isothermal <ideal gas>, $P=\rho c_s^2$ and $c_s$ is independent of $z$, giving
$$
c_s^2\frac{d\rho}{dz}=-\Omega^2\rho z,\qquad
\boxed{\rho(r,z)=\rho_0(r)e^{-z^2/(2h^2)},\quad h=\frac{c_s}{\Omega},\quad\frac hr=\frac{c_s}{v_K}.}
$$
Here $v_K=r\Omega=\sqrt{GM_\star/r}$, $c_s^2=k_BT/(\mu m_p)$ is the isothermal <sound speed> squared, and $h$ is the Gaussian <scale height>. A different convention writing $e^{-z^2/h^2}$ changes the quoted height by $\sqrt2$ but not the physics.
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