Solution: Note that if $g \in G$ such that $\text{ord}(g) = 7 $, then the subgroup generated by $g$, $\{ 1,g,g^2, \dots,g^6 \} $ has order $7$. Also, since $7$ is prime, the order of each $g^i$, for $i =1, 2, \dots,6$ will be $7$. Therefore, each subgroup of $G$ of order $7$ contains exactly $6$ distinct elements of order $7$. Since $G$ has exactly $36$ elements of order $7$, the number of subgroups of $G$ of order $7$ is $\frac{36}{6}= 6$.