jmc

The common ratio between consecutive terms is $\frac{6}{4}=\frac{3}{2}$ (we could have chosen any two consecutive terms and divided the second one by the first to find the common ratio; we chose 4 and 6 because they seemed simple). So the $n^{\text{th}}$ term of the sequence is $\frac{16}{9}\cdot {\left(\frac{3}{2}\right)}^{n-1}$. Plugging in $n=8$, we get $$\frac{16}{9}\cdot {\left(\frac{3}{2}\right)}^7=\frac{2^4}{3^2}\cdot \frac{3^7}{2^7}=\frac{3^5}{2^3}=\boxed{\frac{243}{8}}.$$