Note that $a$ has order $6$ iff $a$ is a solution of the congruence $x^6-1\equiv 0\pmod{p}$, but does not have order $\lt 6$. We can factor $x^6-1$ as $x^6-1=(x^3-1)(x^3+1)=(x^2-1)(x^2+x+1)(x^2-x+1).$ If $a^3-1\equiv 0\pmod p$ or $a+1\equiv 0\pmod{p}$, then $a$ has order $\lt 6$. So any element of order $6$ must be a solution of the congruence $x^2-x+1\equiv 0\pmod{p}$.
Conversely, let $p$ be a prime $\ge 7$. Suppose that $a^2-a+1\equiv 0\pmod p$. It is clear that $a\not\equiv 1\pmod{p}$. And we cannot have $a\equiv -1\pmod{p}$ unless $(-1)^2-(-1)+1\equiv 0\pmod{p}$, which forces $p=3$. And we cannot have $a^2+a+1\equiv 0\pmod{p}$, for that would imply that $(a^2+a+1)-(a^2-a+1)\equiv 0\pmod{p}$. This is impossible, since $2a\equiv 0\pmod{p}$ implies that $a\equiv 0\pmod{p}$, and thus that $a^2-a+1\not\equiv 0\pmod{p}$.
So for $p \ge 7$, the elements of order $6$ are precisely the solutions of the congruence $x^2-x+1\equiv 0\pmod p$. Suppose that $a$ is such a solution. Then $(1-a)^2-(1-a)+1=(1-2a+a^2)-(1-a)+1=a^2-a+1\equiv 0\pmod{p}$, and therefore $1-a$ is also a solution of $x^2-x+1\equiv 0\pmod{p}$. (The congruence has no more than $2$ solutions.)