Answer
$F = \frac{GMm~x}{(R^2+x^2)^{3/2}}$
Work Step by Step
Note that the distance between the mass in the ring and the particle is $\sqrt{R^2+x^2}$
Let $\theta$ be the angle between the central axis $x$ and the displacement vector from the ring to the particle.
We can find the gravitational force on the particle:
$F = \frac{GMm}{(\sqrt{R^2+x^2})^2}~cos~\theta$
$F = \frac{GMm}{R^2+x^2}~\frac{x}{\sqrt{R^2+x^2}}$
$F = \frac{GMm~x}{(R^2+x^2)^{3/2}}$
