1. **Problem Statement:** We are given that Jamie has a probability of $\frac{1}{3}$ to hit a target on any attempt. 2. The probability that he hits the target for the first time on his $n^{th}$ attempt is given as $\frac{64}{2187}$. 3. We want to find the value of $n$. 4. **Formula Used:** The probability that the first success occurs on the $n^{th}$ trial in a sequence of independent trials with success probability $p$ is given by the geometric distribution: $$P(X=n) = (1-p)^{n-1} p$$ where $p = \frac{1}{3}$ is the probability of hitting the target. 5. Substitute $p = \frac{1}{3}$ and $P(X=n) = \frac{64}{2187}$: $$\left(1 - \frac{1}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ 6. Simplify $1 - \frac{1}{3} = \frac{2}{3}$: $$\left(\frac{2}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ 7. Multiply both sides by 3: $$\left(\frac{2}{3}\right)^{n-1} = \frac{64}{2187} \times 3 = \frac{192}{2187}$$ 8. Note that $64 = 2^6$ and $2187 = 3^7$, so: $$\frac{192}{2187} = \frac{64 \times 3}{3^7} = \frac{2^6 \times 3}{3^7} = 2^6 \times 3^{1-7} = 2^6 \times 3^{-6}$$ 9. Also, rewrite the left side: $$\left(\frac{2}{3}\right)^{n-1} = 2^{n-1} \times 3^{-(n-1)}$$ 10. Equate the powers of 2 and 3: $$2^{n-1} \times 3^{-(n-1)} = 2^6 \times 3^{-6}$$ 11. This implies: $$n - 1 = 6$$ and $$-(n - 1) = -6$$ 12. Solve for $n$: $$n = 7$$ **Final answer:** $n = 7$