To find the ratio of the magnetic field at $$\frac{a}{2}$$ and $$2a$$ distances from the axis of a long straight wire, we use Ampère's Law, which relates the magnetic field around a current-carrying conductor to the current enclosed by it.

Ampère’s Law is given by:

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$$

where:

$$\vec{B}$$ is the magnetic field,

$$\mu_0$$ is the permeability of free space,

$$I_{\text{enc}}$$ is the enclosed current.

For a point inside the wire (at radius $$r=a/2$$):

The current enclosed by a radius $$r$$ is proportional to the area of the cross-section at radius $$r$$.

The area of the cross-section at radius $$r$$ is given by:

$$\pi \left( \frac{a}{2} \right)^2 = \frac{\pi a^2}{4}$$

The total current $$I$$ is uniformly distributed, thus the current enclosed $$I_{\text{enc}}$$ at radius $$r = \frac{a}{2}$$ is:

$$I_{\text{enc}} = I \times \frac{\text{Area enclosed}}{\text{Total area}} = I \times \frac{\frac{\pi a^2}{4}}{\pi a^2} = \frac{I}{4}$$

Applying Ampère’s Law inside the conductor, we get:

$$B \cdot 2 \pi \left( \frac{a}{2} \right) = \mu_0 \left( \frac{I}{4} \right)$$

So,

$$B \cdot \pi a = \frac{\mu_0 I}{4}$$

Therefore, the magnetic field inside the wire at $$r = \frac{a}{2}$$ is:

$$B_{\frac{a}{2}} = \frac{\mu_0 I}{4 \pi a}$$

For a point outside the wire (at radius $$r = 2a$$):

The total current enclosed by a radius $$r = 2a$$ is the entire current $$I$$.

Applying Ampère’s Law outside the conductor, we get:

$$B \cdot 2 \pi (2a) = \mu_0 I$$

So,

$$B \cdot 4 \pi a = \mu_0 I$$

Therefore, the magnetic field outside the wire at $$r = 2a$$ is:

$$B_{2a} = \frac{\mu_0 I}{4 \pi a}$$

Hence, the ratio of the magnetic field at $$\frac{a}{2}$$ and $$2a$$ is:

$$\frac{B_{\frac{a}{2}}}{B_{2a}} = \frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = 1:1$$

So, the correct option is:

Option C: $$1:1$$