To solve this problem, we will use the concept of radioactive decay and the formula for half-life.$\\$Given:$\\$• Half-life of isotope $X = 1.4 \times 10^9$ years$\\$• Ratio of $X$ to $Y$ is $1:7 \\$Let $N_0$ be the initial quantity of $X,$ and $N$ be the remaining quantity of $X$ after time $t. \\$Since the ratio of $X$ to $Y$ is $1:7,$ we have:$\\ \frac{N}{N_0 - N} = \frac{1}{7} \\$From this, we can express $N_0$ in terms of $N: \\ N_0 = N + 7N = 8N \\$The formula for radioactive decay is:$\\ N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \\$Substitute $N_0 = 8N$ into the decay formula:$\\ N = 8N \left(\frac{1}{2}\right)^{\frac{t}{1.4 \times 10^9}} \\$Divide both sides by $N: \\ 1 = 8 \left(\frac{1}{2}\right)^{\frac{t}{1.4 \times 10^9}} \\$Rearrange to solve for $t: \\ \left(\frac{1}{2}\right)^{\frac{t}{1.4 \times 10^9}} = \frac{1}{8} \\$Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3,$ we equate the exponents:$\\ \frac{t}{1.4 \times 10^9} = 3 \\$Solve for $t: \\ t = 3 \times 1.4 \times 10^9 \\ t = 4.2 \times 10^9$ years$\\$Therefore, the age of the rock is $4.2 \times 10^9$ years, which corresponds to Option 3.