10. 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 $r>1$. In such a sequence, the smallest value that $\dot{a}_{4}$ can take is $\qquad$ .
10. 27.

Brief solution: Given $r \in \mathbf{Q}$, there exist coprime positive integers $p, q(p>q \geqslant 2)$ such that $r=\frac{q}{p}$, then $a_{4}=a_{1} r^{3}=$ $\frac{a_{1} q^{3}}{p^{3}}$
Since $a_{4} \in \mathbf{Z}$, $a_{1}$ must be a multiple of $p^{3}$. Therefore, we can set $a_{1}=k p^{3}$ ($k$ is a positive integer), thus $a_{4}=k q^{3}$. Since $q>p \geqslant 2$, to make $a_{4}$ the smallest, we must have $k=1, q=3$, at this point $a_{4}=27$, and the sequence is $8,12,18$, 27.