The activity of a radioactive material is $$6.4 \times 10^{-4}$$ curie. Its half life is 5 days. The activity will become $$5 \times 10^{-6}$$ curie after

We are given:

Initial activity: $$A_0 = 6.4 \times 10^{-4}$$ curie

Final activity: $$A = 5 \times 10^{-6}$$ curie

Half-life: $$T_{1/2} = 5$$ days

Using the radioactive decay formula for activity, $$A = A_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$$.

The ratio of activities is $$\frac{A_0}{A} = \frac{6.4 \times 10^{-4}}{5 \times 10^{-6}} = \frac{6.4}{0.05} = 128$$, and since $$128 = 2^7$$, it follows that $$2^{t/T_{1/2}} = 2^7$$ and therefore $$\frac{t}{T_{1/2}} = 7$$.

Hence, $$t = 7 \times T_{1/2} = 7 \times 5 = 35 \text{ days}$$.

The correct answer is Option D: 35 days.