A homogeneous bar of length $L$ and mass $M$ is at a distance h from a point mass $m$ as shown. The force on $m$ is $F$. Then

Consider a small element dx of the rod at a distance x  from the left end of the rod.

Mass of this small element is  $dM=\frac{M}{L}dx,$

Distance of this small element from mass $m$ is  $x+h$

Force on  m due to this small element is  $dF=\frac{GmdM}{(x+h{)}^2}$

To get the total force $F$ on  the mass $m$, we integrate  $dF$ from $x=0$ to $x=L$

$F={\int }_{x=0}^{x=L}\frac{Gm\frac{M}{L}dx}{(x+h{)}^2}=\frac{-GmM}{L}(\frac{1}{(L+h)}-\frac{1}{h})=\frac{GMm}{h(L+h)}$

So, the force on m due to homogeneous bas is $=\frac{GMm}{h(L+h)}.$