I believe we have to check all values of $k$.
$k = 0$: $a^0 = 1 = b^2$. This holds for all $a\in \mathbb{F}_{27}$, so we just have to find all $b \in \mathbb{F}_{27}$ for which $b^2 \equiv 1 \text{ (mod } 3)$. For $b \not\equiv 0 \text{ (mod } 3)$ this congruence holds (As a result of Fermat's Little Theorem). So $b$ can take $18$ values (not $0,3,6,9,12,15,18,21,24$). This means that for $k=0$, we have $27 \cdot 18$ ordered pairs.
$k = 1$: $a = 1 = b^2$. Now we only have $1 \cdot 18$ ordered pairs.
$k = 2$: $a^2 = 1 = b^2$. Same reasoning for the possible values of $a$: $a^2 \equiv 1 \operatorname{ (mod } 3)$ if $a \not\equiv 0 \operatorname{ (mod } 3)$. Now we have $18 \cdot 18$ ordered pairs.
$k = 3$: $a^3 = 1 = b^2$. Now $a^3 \equiv 1 \operatorname{ (mod } 3)$, so $a \equiv 1 \operatorname{ (mod } 3)$. There are $9 \cdot 18$ ordered pairs.
Now for $k \ge 4$: $a^k = 1 = b^2$. The calculation of the values of $a$ can be reduced to one of the prior situations, assuming that $a$ is not a multiple of 3 (and - of course - neither is $b$).
So the cases $k=1, k=2, k=3$ can be rewritten as $k \equiv 1, k\equiv 2, k \equiv 0 (k \ne 0) \operatorname{ (mod } 3)$.
Is my method correct?