Answer
The acceleration of the ball at P is initially $a= -2.603\times 10^{-9}\mathrm{ms^{-2}}$. It acts in the downward direction.
Work Step by Step
Let the angle between the line joining the A and B and the line joining A and P(or B and P) be $\theta$. From the diagram, $\sin\theta = 6/8 = 0.75$.
From the diagram, it is clear that the force on the ball at P due to the balls at A and B are in the directions $\vec{PA}$ and $\vec{PB}$ respectively. Since the mass of the fixed balls at A and B are equal, the magnitude of the forces is equal. However, from the diagram, it is clear that the component of force on the ball at P parallel(and anti-parallel) to $\vec{AB}$ from the two balls are equal and in opposite directions and thus they cancel. The only component of the two forces that remains is the component perpendicular to $\vec{AB}$ which is in the downwards direction and is given by
$F = -\frac{2GMm}{r^2}\sin\theta$,
where $M = 0.260\mathrm{kg}$, $m = 0.010\mathrm{kg}$ and $r = 10\mathrm{cm} = 0.1\mathrm{m}$ and the factor of 2 comes from taking into account the force due to both the balls.
Thus the acceleration of the ball at P is
$a = F/m = -\frac{2GM}{r^2}\sin\theta = -\frac{2\times 6.674\times10^{-11}\mathrm{m^3kg^{-1}s^{-2}}\times0.260\mathrm{kg}}{(0.1\mathrm{m})^2}\times 0.75$
$a = -2.603\times 10^{-9}\mathrm{ms^{-2}}$
