To solve this problem, we need to determine the time it takes for the activity of a radioactive substance to drop to $\frac{1}{16}$ of its initial value.$\\ \\$The half-life of the substance is given as 20 minutes.$\\ \\$The activity of a radioactive substance decreases exponentially over time, and the relationship is given by:$\\ A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \\$where:$\\$• $A$ is the final activity,$\\$• $A_0$ is the initial activity,$\\$• $T_{1/2}$ is the half-life, and$\\$• $t$ is the time elapsed.$\\ \\$We want the activity to drop to $\frac{1}{16}$ of its initial value, so:$\\ A = \frac{A_0}{16} \\ \\$Substitute this into the exponential decay formula:$\\ \frac{A_0}{16} = A_0 \left(\frac{1}{2}\right)^{\frac{t}{20}} \\ \\$Cancel $A_0$ from both sides:$\\ \frac{1}{16} = \left(\frac{1}{2}\right)^{\frac{t}{20}} \\ \\$We know that $\frac{1}{16} = \left(\frac{1}{2}\right)^4,$ so:$\\ \left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^{\frac{t}{20}} \\ \\$Since the bases are the same, equate the exponents:$\\ 4 = \frac{t}{20} \\ \\$Solve for $t: \\ t = 4 \times 20 \\ t = 80$ minutes$\\ \\$Therefore, the time required for the activity to drop to $\frac{1}{16}$ of its initial value is 80 minutes.$\\$This corresponds to Option 4.