Step 1: Determine the number of half-lives ($n$) that have passed. The fraction of radioactive atoms remaining after time $t$ is given by the formula: $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$$ where $N$ is the number of atoms remaining, $N_0$ is the initial number of atoms, and $n$ is the number of half-lives. Given that the fraction of atoms left after 120 years is $\frac{1}{64}$: $$\frac{1}{64} = \left(\frac{1}{2}\right)^n$$ We know that $2^6 = 64$, so $\frac{1}{64} = \left(\frac{1}{2}\right)^6$. Therefore, $n = 6$. This means 6 half-lives have passed in 120 years. Step 2: Calculate the half-life ($T_{1/2}$) of the material. The number of half-lives ($n$) is also related to the total time ($t$) and the half-life ($T_{1/2}$) by the formula: $$n = \frac{t}{T_{1/2}}$$ We need to find $T_{1/2}$, so we can rearrange the formula: $$T_{1/2} = \frac{t}{n}$$ Substitute the given total time $t = 120$ years and the calculated number of half-lives $n = 6$: $$T_{1/2} = \frac{120 \, \text{years}}{6}$$ $$T_{1/2} = 20 \, \text{years}$$ Step 3: Compare the result with the given options. The calculated half-life is 20 years, which corresponds to option C. The half-life of the material is $\boxed{\text{20 years}}$.