Solution:

We observe that $54 = 2 \cdot 3^{3}$, and therefore $a$ is divisible by both $2$ and $3$, and has no prime factors other than $2$ and $3$. In other words, $a$ can be written in the form $a = 2^{x} \cdot 3^{y}$ for suitable positive integers $x$ and $y$. It follows that
$$54^{a} = \left(2 \cdot 3^{3}\right)^{2^{x} \cdot 3^{y}} = 2^{2^{x} \cdot 3^{y}} \cdot 3^{3 \cdot 2^{x} \cdot 3^{y}} \quad \text{and} \quad a^{b} = \left(2^{x} \cdot 3^{y}\right)^{b} = 2^{x b} \cdot 3^{y b}$$
By equating the exponents of $2$ and of $3$ in the two expressions we deduce that
$$2^{x} \cdot 3^{y} = x b \quad \text{and} \quad 3 \cdot 2^{x} \cdot 3^{y} = y b.$$
Comparing the two equalities we conclude that $y = 3x$, and consequently
$$a = 2^{x} \cdot 3^{y} = 2^{x} \cdot 3^{3x} = \left(2 \cdot 3^{3}\right)^{x} = 54^{x},$$
as required.