To determine the acceleration of a cylinder rolling down an inclined plane without slipping, we can use Newton's second law and the concept of rolling motion. For an inclined plane at an angle $$ \theta $$, the component of gravitational acceleration along the plane is $$ g \sin \theta $$. However, because the cylinder is rolling and not sliding, not all of this component accelerates the center of mass; some of it goes into causing rotational acceleration about the center of mass.

For a rolling cylinder, the moment of inertia $$ I $$ is $$ I = \frac{1}{2} m r^2 $$, where $$ m $$ is the mass of the cylinder and $$ r $$ is the radius. The condition for rolling without slipping is that the linear acceleration $$ a $$ of the center of mass is equal to the radius $$ r $$ times the angular acceleration $$ \alpha $$, i.e., $$ a = r \alpha $$.

To find the linear acceleration $$ a $$, we use the torque $$ \tau $$ about the center of mass caused by the gravitational force down the incline. The torque due to gravity is $$ \tau = mg \sin \theta \cdot r $$, and from Newton's second law for rotation, the angular acceleration is given by

$$ \alpha = \frac{\tau}{I} = \frac{mg \sin \theta \cdot r}{\frac{1}{2} m r^2} = \frac{2g \sin \theta}{r} $$

Now using $$ a = r \alpha $$:

$$ a = r \left( \frac{2g \sin \theta}{r} \right) = 2g \sin \theta $$

But we must account for the fact that only a portion of the gravitational acceleration goes into translating the cylinder down the plane due to the rolling condition. This is where we apply the concept of the ``rolling factor'' for a cylinder, which is $$ \frac{2}{3} $$ for a solid cylinder, meaning $$ \frac{2}{3} $$ of the gravitational component is used for translation.

The acceleration of the center of mass for the cylinder is therefore:

$$ a = \frac{2}{3} g \sin \theta $$

Now we plug in the values of $$ \theta = 60^{\circ} $$ (which has $$ \sin 60^{\circ} = \frac{\sqrt{3}}{2} $$) and $$ g = 10 \; m/s^2 $$:

$$ a = \frac{2}{3} \cdot 10 \cdot \frac{\sqrt{3}}{2} = \frac{10 \sqrt{3}}{3} $$

To match the given expression $$ \frac{x}{\sqrt{3}} m / s^2 $$, let's manipulate our expression for $$ a $$:

$$ a = \frac{10 \sqrt{3}}{3} = \frac{10 \sqrt{3}}{3} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{10 \cdot 3}{3 \cdot \sqrt{3}} = \frac{10}{\sqrt{3}} m/s^2 $$

Hence, the value of $$ x $$ is $$ 10 $$.