Answer
$4.45 \times 10^{-9} m/s^2$, toward the lighter (8 kg) mass.
Work Step by Step
The particle (of mass $M$) is at a distance of $x_1 = 0.2 m$ from $m_1 = 8 kg$.
Obviously, given the set-up, this means that it is at a distance $x_2 = 0.3 m$ from $m_2$ = 12 kg.
Now, $F_1 = G\frac{m_1M}{x_1^2} = (6.67 \times 10^{-11})\frac{(8)M}{(0.2)^2} = (1.33 \times 10^{-8})M \ N$ and $F_2 = G\frac{m_2M}{x_2^2} = (6.67 \times 10^{-11})\frac{(12)M}{(0.3)^2} = (8.89 \times 10^{-8})M \ N$.
Since $F_1$ and $F_2$ point in opposite directions, the net force $F_{net} = F_1 - F_2 = (4.45\times 10^{-9})M \ N$. (Here we've arbitrarily chosen the direction of the lighter mass to be the +x-direction.)
This is of the form $F = Ma$ for some acceleration $a$. Therefore, dividing by the mass $M$, we find the acceleration to be $4.45 \times 10^{-9} m/s^2$, toward the lighter mass.
