The half life period of a radioactive substance is 60 days. The time taken for $$\frac{7}{8}$$th of its original mass to disintegrate will be:

We are given that the half-life of a radioactive substance is $$T_{1/2} = 60$$ days and we need to find the time taken for $$\frac{7}{8}$$th of the original mass to disintegrate.

If $$\frac{7}{8}$$th of the mass has disintegrated, the remaining mass is:

$$\frac{N}{N_0} = 1 - \frac{7}{8} = \frac{1}{8}$$

Using the radioactive decay formula:

$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/T_{1/2}}$$

Substituting the values:

$$\frac{1}{8} = \left(\frac{1}{2}\right)^{t/60}$$

Since $$\frac{1}{8} = \left(\frac{1}{2}\right)^3$$, we get:

$$\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{t/60}$$

Comparing the exponents:

$$\frac{t}{60} = 3$$

$$t = 180 \text{ days}$$

Therefore, the correct answer is Option C: 180 days.