I've show that $\frac{9}{8}$ is an upper bound, but am struggling with showing that it is the least upper bound, or rather how $\frac{a^2}{2^a} > \frac{9}{8} - \epsilon$ shows that $\frac{9}{8}$ is a least upper bound. My proof as is:
We first prove the inequality $2^n \geq n^2$ for $n \geq 4$. We do so by mathematical induction. For the basis step let $n = 4$. Then \begin{align*} 2^{(4)} &\geq (4)^2 \\ 16 &\geq 16. \end{align*}
Thus, $n = 4$ holds true. Now assume that the statement is true for some $n = k \geq 4$. That is, $k^2 \leq 2^k$. This implies that $n = k+1$ is true. Then
\begin{align*} 2^{k+1} &= 2 \cdot 2^k \\ &\geq 2 \cdot k^2 \tag{By Induction Hypothesis}\\ & \geq k^2 + 2k + 1 \\ & \geq (k+1)^2. \end{align*}
Hence, $2^{k+1} \geq (k+1)^2$. Therefore, it follows that $2^n \geq n^2$. Thus, when $n=3$ we have that $\frac{9}{8}$ is an upper bound of $A$. Since $\frac{9}{8}$ is an upper bound of A we have for any $a \in A$ that $a \leq \frac{9}{8}$. Now, suppose sup$(A) = \frac{9}{8}$. Then, for any $\epsilon > 0, \frac{9}{8} - \epsilon$ is not an upper bound of $A$. Hence, $\frac{a^2}{2^a} > \frac{9}{8} - \epsilon$.