Given that the fraction of atoms left after 120 years is (1)/(64):

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.

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 $n = 6$:

$$T_{1/2} = \frac{120 \, \text{years}}{6}$$ $$T_{1/2} = 20 \, \text{years}$$

The calculated half-life is 20 years, which corresponds to option C.

The half-life of the material is $\text{20 years}$.