Object A increases in mass, and so increases in volume
I'm going to make the assumption that we are adding mass to A by providing more material of the same density $\rho_{A}$, rather than exchanging the current material with denser material or adding varying densities. I don't have to make that assumption, but it seems like it's what you're going for and we'll see where it leads. I'll also let A and B be spherical, to make them planet-like and also stable, so we don't have to worry about their shape changing. (assume a spherical cow, amirite?)
Once A is large enough to touch B, the distance between their centers is $r_{A}+r_{B}$. The mass of A is $\rho_{A}(\frac{4}{3} \pi r_{A}^{3})$. So the gravitational force between them is:
\begin{equation} F=\frac{G \rho_{A} (\frac{4}{3} \pi r_{A}^{3})m_{B}}{(r_{a}+r_{b})^{2}} \end{equation}
It's clear that the top will grow more quickly than the bottom (everything is constant except for the radii). 15
After a certain point, a cube grows faster than a square. Thus, the force will continue to increase, almost linearly. But will A turn into a black hole?
The critical point at which A turns into a black hole is when the following equation is satisfied:
\begin{equation} \frac{m_{A}}{r_{A}}=\frac{c^{2}}{2G} \end{equation}
c is the speed of light, G is the gravitational constant as above.
We also know:
\begin{equation} \frac{m_{A}}{r_{A}}=\rho_{A}\frac{4\pi r^{2}}{3} \end{equation}
(This comes from $V_{A}=\frac{4}{3}\pi r_{A}^{3}$)
Thus, the black hole happens when:
\begin{equation} r_{A}=\sqrt{\frac{3c^{2}}{8\pi \rho G}}=\frac{1.27*10^{13}}{\sqrt{\rho_{A}}} \end{equation}
So the gravity on B increases almost linearly until a black hole is created. The gravity then only keeps increasing (of course, B would be a part of A at this point too). If we relaxed our assumption of constant density of A, the gravitational force would only increase. There is no sweet spot. 483