Answer
The maximum gravitational force between the balls has a magnitude of $9.58\times10^{-9}N$
Work Step by Step
The gravitational force between the bowling ball with mass $m_{bo}$ and the billiard ball with mass $m_{bi}$ (the distance between two balls is $r$) is $$F_g=G\frac{m_{bo}m_{bi}}{r^2}$$
From the formula, $F_g$ attains its maximum value when the distance between the balls $r$ is minimum, or when the balls are next to each other, only separated by the radius of each ball.
So $r_{min}$ is the sum of the radii of these balls. In other words, $r_{min}=0.11+0.028=0.138m$
We have $m_{bo}=7.2kg$, $m_{bi}=0.38kg$, $G=6.67\times10^{-11}Nm^2/kg^2$
$$F_{g, max}=G\frac{m_{bo}m_{bi}}{r_{min}^2}=9.58\times10^{-9}N$$
