The correct answer is $\boxed{\frac{10}{21}}$.
The probability of getting at least 6 on two rolls of a six-sided die is the sum of the probabilities of getting 6 on the first roll and 5 or more on the second roll, getting 5 on the first roll and 6 or more on the second roll, and getting 4 on the first roll and 7 or more on the second roll.
The probability of getting 6 on the first roll is $\frac{1}{6}$. The probability of getting 5 or more on the second roll is $\frac{5}{6}$, since there are 5 possible outcomes that satisfy this condition (namely, 5, 6, 7, 8, and 9). The probability of getting 5 on the first roll and 6 or more on the second roll is $\frac{1}{6} \cdot \frac{5}{6} = \frac{5}{36}$. The probability of getting 4 on the first roll and 7 or more on the second roll is $\frac{1}{6} \cdot \frac{5}{6} = \frac{5}{36}$.
Therefore, the probability of getting at least 6 on two rolls of a six-sided die is $\frac{1}{6} + \frac{5}{36} + \frac{5}{36} = \boxed{\frac{10}{21}}$.
Option A is incorrect because it is the probability of getting at least 6 on a single roll of a six-sided die. Option B is incorrect because it is the probability of getting 5 or more on a single roll of a six-sided die. Option C is incorrect because it is the probability of getting 2 or more on a single roll of a six-sided die. Option D is incorrect because it is the probability of getting 1 on a single roll of a six-sided die.