The relationship between the root-mean-square (rms) speed ($$v_{rms}$$) and the average speed ($$v_{avg}$$) of molecules in a gas can be found using the Maxwell-Boltzmann distribution. The rms speed and average speed are related as follows:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$ $$v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$

Where:

• $$R$$ is the ideal gas constant
• $$T$$ is the temperature in Kelvin
• $$M$$ is the molar mass of the gas
• $$\pi$$ is the mathematical constant pi

In this problem, the rms speed of the oxygen molecule is given by:

$$v_{rms} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}} v_{avg}$$

Now, let's divide the expression for $$v_{rms}$$ by the expression for $$v_{avg}$$:

$$\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}}$$

By simplifying the expression, we get:

$$\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{3}}{\sqrt{\frac{8}{\pi}}} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}}$$

Square both sides of the equation:

$$\frac{3}{\frac{8}{\pi}} = 1 + \frac{5}{x}$$

Now we will substitute the provided value of $$\pi = \frac{22}{7}$$:

$$\frac{3}{\frac{8}{\frac{22}{7}}} = 1 + \frac{5}{x}$$

By simplifying the expression, we get:

$$\frac{3 \cdot \frac{22}{7}}{8} = 1 + \frac{5}{x}$$

Now let's solve for $$x$$:

$$\frac{66}{56} - 1 = \frac{5}{x}$$ $$\frac{10}{56} = \frac{5}{x}$$

Multiplying both sides by $$x$$:

$$\frac{10}{56}x = 5$$

Finally, solving for $$x$$:

$$x = \frac{5 \cdot 56}{10} = 28$$

So, the value of $$x$$ is $$\boxed{28}$$.