I'll assume that an ordered pair $(b, e)$ means that $b \leqslant e$. If it in fact means $b \geqslant e$, just draw the Hasse diagram I'm describing and turn it upside down :-)
To start, I'll just make a quick table of who's "less than" whom.
\begin{array}{l|l}
a & f\\
b & d, e\\
c & a, f\\
d & \\
e & \\
f & \\
\end{array}
where the $b$ row corresponds to the fact that $b \leqslant d$ and $b \leqslant e$. Because partial orders are reflexive, I haven't bothered to list $x \leqslant x$, since we know that to be the case and displaying it only makes the relevant information less visible.
Taking $d$ as an example, if $d \leq y$, then $y = d$; there's nothing above $d$ in the Hasse diagram. All of $d, e$, and $f$ are in top of the Hasse diagram; they're never below anything. A poset can have several maximal elements, and they need not be on the same "level" (and that's the case here).
Oppositely, take $b$. If $x \leqslant b$, then $x = b$; nothing is below $b$ (or $c$) in the Hasse diagram. They're the minimal elements, and are drawn at the same "level."
That leaves only $a$, who is both above $c$, and below $f$; we have $c \leqslant a \leqslant f$.
Try and see if you can work something out here. Below I'll list a couple more hints, that should verify if you're right or not (but don't look until you try!).
The Hasse diagram here will be disconnected. One piece will just be a straight "line" for the $c \leqslant a \leqslant f$ (with $c$ at the bottom). The other piece will be a kind of V shape, with $b$ at the bottom and $e, d$ at the top.
You missed the edges 24-72 and 4-36.
$\inf_A\{16,18\}$, if it exists, is the greatest lower bound of both 16 and 18. The lower bounds of 16 are $\{2,4,8\}$ and the lower bounds of 18 are $\{2,6\}$. 2 is the only common lower bound so it is the greatest and $\inf_A\{16,18\} = 2$.
A similar effort should show you that $\sup_A\{4,6\} = 24$.
Best Answer
The diagram looks right.
But I don't agree with "the least upper bound of $\{2\}$ is $4$". Namely, $2$ is another, smaller upper bound.
Also, under the "divides" relation, 12 and 5 are incomparable, so neither of them can be a greatest element.