According to Bohr's quantization of angular momentum, the angular momentum $$L$$ of a particle in a circular orbit is given by:

$$L = n\hbar$$

Where $$n$$ is an integer and $$\hbar$$ is the reduced Planck's constant. The angular momentum $$L$$ can also be expressed as:

$$L = mvr$$

Where $$m$$ is the mass of the particle, $$v$$ is its linear velocity, and $$r$$ is the radius of the orbit.

Now, we are given the potential energy $$U = \frac{1}{2} m\omega^2r^2$$. Since the particle is in a circular orbit, its centripetal force is provided by the gradient of the potential energy:

$$m\frac{v^2}{r} = -\frac{\mathrm{d}U}{\mathrm{d}r} = -m\omega^2r$$

We can simplify this equation to get the relation between $$v$$ and $$r$$:

$$v^2 = \omega^2r^2$$

Now, let's combine the equations for angular momentum and the relation between $$v$$ and $$r$$:

$$n\hbar = mvr = m\sqrt{\omega^2r^2}r = m\omega r^2$$

We can now solve for the radius $$r$$ in terms of $$n$$:

$$r^2 = \frac{n\hbar}{m\omega}$$

Taking the square root of both sides, we get:

$$r \propto \sqrt{n}$$

So, the radius of the $$n^{\text{th}}$$ orbit is proportional to $$\sqrt{n}$$.