Answer
$a_n=\frac{32}{27}\cdot(\frac{3\sqrt[3]{12}}{4})^{n-1}$
Work Step by Step
The nth term of a geometric series can be obtained by the following formula: $a_n=a_1\cdot r^{n-1}$, where $a_1$ is the first term and $r$ is the common ratio.
Hence here: $a_7/a_4=r^3$, thus $r=\sqrt[3]{\dfrac{\frac{243}{8}}{6}}=\frac{3\sqrt[3]{12}}{4}$ and $a_4=a_1r^3$
This gives:
$a_1=\frac{a_4}{r^3}=\frac{32}{27}$, hence $a_n=\frac{32}{27}\cdot(\frac{3\sqrt[3]{12}}{4})^{n-1}$
