For a first-order reaction, the integrated rate law is given by:

$$\ln \frac{[A]}{[A]_0} = -kt$$

where:

$$[A]_0$$ is the initial concentration,

$$[A]$$ is the concentration at time $$t$$,

$$k$$ is the rate constant.

Let's find the times $$t_1$$ and $$t_2$$ corresponding to when the concentration becomes $$\frac{1}{4}[A]_0$$ and $$\frac{1}{8}[A]_0$$ respectively.

For concentration $$\frac{1}{4}[A]_0$$:

Substitute into the integrated rate law:

$$\ln\left(\frac{1}{4}\right) = -kt_1$$

Recognizing that $$\ln\left(\frac{1}{4}\right) = -\ln 4$$, we have:

$$-\ln 4 = -kt_1 \quad \Longrightarrow \quad t_1 = \frac{\ln 4}{k}$$

For concentration $$\frac{1}{8}[A]_0$$:

Substitute into the integrated rate law:

$$\ln\left(\frac{1}{8}\right) = -kt_2$$

Recognizing that $$\ln\left(\frac{1}{8}\right) = -\ln 8$$, we have:

$$-\ln 8 = -kt_2 \quad \Longrightarrow \quad t_2 = \frac{\ln 8}{k}$$

Now, to find the ratio $$\frac{t_1}{t_2}$$:

$$ \frac{t_1}{t_2} = \frac{\frac{\ln 4}{k}}{\frac{\ln 8}{k}} = \frac{\ln 4}{\ln 8} $$

We can simplify further by expressing the logarithms in terms of $$\ln 2$$:

$$\ln 4 = \ln (2^2) = 2\ln 2$$

$$\ln 8 = \ln (2^3) = 3\ln 2$$

Thus,

$$ \frac{t_1}{t_2} = \frac{2\ln 2}{3\ln 2} = \frac{2}{3} $$

So, the ratio $$\frac{t_1}{t_2}$$ is $$\frac{2}{3}$$.

This corresponds to Option D.