= Solution
Let the uniform thin <galactic disk> have outer radius $R_d$ and <surface mass density> $\Sigma=M/(\pi R_d^2)$, where $M$ is the <mass> calculated in <6/a>. If the redistribution retains the stated outer radius, set $R_d=16\ \mathrm{kpc}$; otherwise $M$ alone does not determine both $R_d$ and $\Sigma$.
A ring of radius $s$ and width $ds$ has <mass> $dM=2\pi\Sigma s\,ds$. By axial symmetry its transverse gravitational force components cancel. At signed height $z$, its vertical force on a <star> of <mass> $m$ is
$$
dF_z=-\frac{Gm\,z\,dM}{(s^2+z^2)^{3/2}}.
$$
Integrating the rings gives, for $z\ne0$,
$$
\begin{aligned}
F_z&=-2\pi Gm\Sigma z\int_0^{R_d}\frac{s\,ds}{(s^2+z^2)^{3/2}}\\
&=-2\pi Gm\Sigma z\left(\frac1{|z|}-\frac1{\sqrt{z^2+R_d^2}}\right).
\end{aligned}
$$
Thus
$$
\boxed{\mathbf F(z)=-2\pi Gm\Sigma\left[\operatorname{sgn}z-\frac z{\sqrt{z^2+R_d^2}}\right]\hat{\mathbf z}.}
$$
For a positive distance above the plane, this is $\mathbf F=-2\pi Gm\Sigma[1-z/\sqrt{z^2+R_d^2}]\hat{\mathbf z}$, directed toward the disk. With $R_d=16\ \mathrm{kpc}$, the inferred $\Sigma\simeq178.3M_\odot\,\mathrm{pc^{-2}}$ sets its normalization.
The far-field expansion gives $F_z\simeq-GMm\operatorname{sgn}(z)/z^2$, as required for a point-like total <mass>. Close to either face, $F_z\to\mp2\pi Gm\Sigma$, the infinite-sheet limit. The ideal zero-thickness model has a jump in gravitational field at $z=0$; the formula is understood through its one-sided limits. Its symmetric mean is zero on the plane, and a disk of finite thickness resolves the discontinuity.
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