The activity of a radioactive material is $$2.56 \times 10^{-3}$$ Ci. If the half life of the material is 5 days, after how many days the activity will become $$2 \times 10^{-5}$$ Ci?

We are given the initial activity $$A_0 = 2.56 \times 10^{-3}$$ Ci and the final activity $$A = 2 \times 10^{-5}$$ Ci, with a half-life $$t_{1/2} = 5$$ days.

The radioactive decay law gives:

$$A = A_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}$$

Substituting the values:

$$2 \times 10^{-5} = 2.56 \times 10^{-3} \times \left(\frac{1}{2}\right)^{t/5}$$

Dividing both sides by $$2.56 \times 10^{-3}$$:

$$\frac{2 \times 10^{-5}}{2.56 \times 10^{-3}} = \left(\frac{1}{2}\right)^{t/5}$$

$$\frac{2}{256} = \left(\frac{1}{2}\right)^{t/5}$$

$$\frac{1}{128} = \left(\frac{1}{2}\right)^{t/5}$$

Since $$\frac{1}{128} = \left(\frac{1}{2}\right)^7$$, we get:

$$\left(\frac{1}{2}\right)^7 = \left(\frac{1}{2}\right)^{t/5}$$

Therefore:

$$\frac{t}{5} = 7$$

$$t = 35 \text{ days}$$

Hence, the correct answer is Option B (35 days).