jmc

To get from the seventh term to the tenth term, we multiply the seventh term by 3 ($7\times 3=21$). So to get from the tenth term to the $13$th term, we multiply the tenth term by 3 since the ratio between terms is constant. The $13$th term is $21\times 3=\boxed{63}$.

To be more specific, we can write the eighth term as $7r$, where $r$ is the common ratio of the geometric sequence. The ninth term is $7r^2$, the tenth term is $7r^3$, etc. If $7r^3=21$, then $r^3=3$. So $a_n=a_{n-3}{r}^3=3a_{n-3}$. We get the $13$th term with $3a_{10}=3\times 21=\boxed{63}$.