In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $$t_1$$ and $$t_2$$ (s), respectively. The ratio $$t_1/t_2$$ will :

For a first-order reaction, the integrated rate law is
$$k = \frac{2.303}{t}\,\log\!\left(\frac{[R]_0}{[R]}\right)$$
Re-arranging, the time required to reach any concentration is
$$t = \frac{2.303}{k}\,\log\!\left(\frac{[R]_0}{[R]}\right)$$

Concentration falls to one-fourth of the initial value.
Then $$[R] = \frac{[R]_0}{4}$$, so
$$t_1 = \frac{2.303}{k}\,\log\!\left(\frac{[R]_0}{[R]_0/4}\right) = \frac{2.303}{k}\,\log 4$$

Concentration falls to one-eighth of the initial value.
Then $$[R] = \frac{[R]_0}{8}$$, so
$$t_2 = \frac{2.303}{k}\,\log\!\left(\frac{[R]_0}{[R]_0/8}\right) = \frac{2.303}{k}\,\log 8$$

Taking the ratio:
$$\frac{t_1}{t_2} = \frac{\log 4}{\log 8}$$

Using base-10 logarithms:
$$\log 4 = \log(2^2) = 2\log 2$$
$$\log 8 = \log(2^3) = 3\log 2$$

Hence
$$\frac{t_1}{t_2} = \frac{2\log 2}{3\log 2} = \frac{2}{3}$$

Therefore, the required ratio is $$\frac{2}{3}$$, which matches Option D.