$\textbf{Cyclic groups}$ are groups in which every element is a power of some fixed element.
$\textbf{Definition:}$ Let $\text{G}$ be a group, $g \in \text{G}.$ Let $1$ be the identity element. The order of $g$ is the smallest positive integer $n$ such that $g^{n} = 1.$ If there is no positive integer $n$ such that $g^{n} = 1,$ then $g$ has infinite order.
The order of an element is the smallest power which gives the identity the element in two ways. It is smallest in the sense of being numerically smallest, but it is also smallest in the sense that it divides any power which gives the identity.
$\textbf{Definition:}$ If $G$ is a group and $g \in \text{G},$ then the subgroup generated by $g$ is $\langle g \rangle = \{g^{n} | n \in \mathbf{Z}\}.$
$\textbf{Definition:}$ A group $\text{G}$ is cyclic if $\text{G} = \langle g \rangle$ for some $g \in \text{G}.\; g$ is a generator of $\langle g \rangle.$ If a generator $g$ has order $n, \text{G} = \langle g \rangle$ is cyclic of order $n.$ If a generator $g$ has infinite order, $\text{G} = \langle g \rangle$ is infinite cyclic.
Since, $\text{G}$ is a cyclic group with generator ‘$a$’, where Order of ‘$a$’ is $38.$ So, Order of $\text{G}$ is $38.$ Order of an element of the group divides the order of the Group. Assume the order of $a^{32}$ is $m$ then $m$ divides $38.$ So, possible values to check for $m$ is $1$ or $2$ or $19$ or $38.$
And since the order of $a^{32}$ is $m$ then $(a^{32} )^{m} = 1.$
So, $m$ can not be $1$ (because if $m=1$ then order of “$a$” would be $32,$ but it is $38$ in the question)
So, $m$ cannot be $2,$ because if $m=2,$ then $(a^{32} )^{2} = 1,$ so, $a^{64} = 1,$ So, $a^{64} = a^{38}.a^{26} = 1,$ So, $a^{26} = 1,$ then order of “$a$” would be $26,$ but it is $38$ in the question)
$m = 19$ works. $19$ divides $38$ and also $(a^{32} )^{19} = 1.$
Since, order of an element $g$ is smallest positive integer for which power of $g$ becomes identity element, so, $38$ cannot be order $m$ here as $19$ is already working.
$\textbf{Conclusion:}$ In any cyclic group with generator $g,$ order $n;$ order of any element $g^{p}$ will be the smallest positive integer $m$ such that $n$ divides $pm.$
Where $(m,n)$ means $\gcd(m,n).$
Basically find the minimum $m$ for which $(a^{32} )^{m} = 1.$ Also, we know $(a^{38} ) = 1.$
So, $38$ divides $32*m.$ So, $19$ divides $16m.$
We know that “If $p$ is prime and $p|qr ,$ then $p|q$ or $p|r$” .
So, the minimum possible value of $m$ is $19.$
