1. **Problem statement:** We are given a geometric progression (G.P.) where the third term is 10 and the sixth term is 80. We need to find the common ratio $r$. 2. **Formula for the $n$th term of a G.P.:** $$a_n = a_1 r^{n-1}$$ where $a_1$ is the first term and $r$ is the common ratio. 3. **Write equations for the given terms:** - Third term: $$a_3 = a_1 r^{2} = 10$$ - Sixth term: $$a_6 = a_1 r^{5} = 80$$ 4. **Divide the sixth term by the third term to eliminate $a_1$:** $$\frac{a_6}{a_3} = \frac{a_1 r^{5}}{a_1 r^{2}} = r^{3} = \frac{80}{10} = 8$$ 5. **Solve for $r$:** $$r^{3} = 8$$ Taking cube root on both sides: $$r = \sqrt[3]{8} = 2$$ 6. **Answer:** The common ratio is $2$. This corresponds to option A.