The half life of a radioactive substances in 5 days. What percentage of it will remain after 20days?

The half-life ($T_{1/2}$) is 5 days, and the total time ($t$) is 20 days.
The number of half-lives ($n$) is given by:

$$n = \frac{t}{T_{1/2}}$$ $$n = \frac{20 \text{ days}}{5 \text{ days}}$$ $$n = 4$$

After $n$ half-lives, the fraction of the substance remaining ($\frac{N}{N_0}$) is given by:

$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$$ $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^4$$ $$\frac{N}{N_0} = \frac{1}{16}$$

To find the percentage remaining, multiply the fraction by 100%:

$$\text{Percentage remaining} = \frac{1}{16} \times 100\%$$ $$\text{Percentage remaining} = 0.0625 \times 100\%$$ $$\text{Percentage remaining} = 6.25\%$$

Comparing this to the given options:

(a) 12.5%

(b) 25%

(c) 6.25%

(d) 42%

The correct option is (c).

The final answer is $\boxed{\text{6.25\%}}$.