A radioactive isotope has a half-life of 24 minutes. How long will it take for the activity to drop to 1/8 of its initial value?

The activity of a radioactive isotope decreases by half with each passing half-life.
After 1 half-life, the activity is $\frac{1}{2}$ of the initial value.
After 2 half-lives, the activity is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the initial value.
After 3 half-lives, the activity is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ of the initial value.
So, it takes 3 half-lives for the activity to drop to $\frac{1}{8}$ of its initial value.

Alternatively, using the decay formula:

$$\frac{N_t}{N_0} = \left(\frac{1}{2}\right)^n$$

where $N_t$ is the final activity, $N_0$ is the initial activity, and $n$ is the number of half-lives.
We are given $\frac{N_t}{N_0} = \frac{1}{8}$.

$$\frac{1}{8} = \left(\frac{1}{2}\right)^n$$

Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$, we have $n=3$.

The half-life of the isotope is 24 minutes.
Total time = Number of half-lives $\times$ Half-life period

$$\text{Total time} = 3 \times 24 \text{ minutes}$$ $$\text{Total time} = 72 \text{ minutes}$$

The activity will drop to $\frac{1}{8}$ of its initial value in 72 minutes. This corresponds to option C.

The final answer is $\text{C}$.