6. Let the sequence of positive integers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4}$ be a geometric sequence, with the common ratio $r$ not being an integer and $r>1$. The smallest value that $a_{4}$ can take in such a sequence is $\qquad$ .
6. 27 .

According to the problem, $r$ is a rational number, so there exist coprime positive integers $p$ and $q (q > p \geqslant 2)$, such that $r=\frac{q}{p}$. Therefore, $a_{4}=a_{1} r^{3}=\frac{a_{1} q^{3}}{p^{3}}$.
Since $a_{4}$ is an integer, $a_{1}$ must be a multiple of $p^{3}$.
Thus, let $a_{1}=k p^{3}$ ($k$ is a positive integer).
At this point, $a_{4}=k q^{3}$. Note: Given $q > p \geqslant 2$, to make $a_{4}$ the smallest, we have $k=1, q=3$.