Answer = D.
The set $A$ of all words (finite sequence of letters $a−z$), denoted by $W$, in dictionary order.
$W$ is a Poset as it is Reflexive, Antisymmetric and transitive. $W$ is even a Total Ordered structure as Every Two elements of Set $A$ are comparable.
Let's catch the bigger fish here i.e. Well-order. :
A set S together with partial order ≪ is called a well order if it has no infinite descending chains, i.e. there is no infinite sequence x1,x2,..... of elements from S such that xi+1≪ xi and xi+1≠xi for all i.
We know a different definition of Well-ordered set which is "A well-ordered set is a structure of the form $(S, ≤)$ such that $≤$ is a partial order on $S$ and Every nonempty subset of S has a $≤-smallest \,\, element$...We will call this definition as Definition 1 of Well Order. If $(S, ≤)$ is a well-ordered set, we may express this by saying that the relation $≤$ well-orders $S$. (NOTE that We need not say that $S$ first should be a Total ordered set to be a Well Ordered set Because the first and second condition of the definition together itself imply that $(S, \leq)$) is a Total ordered structure)
Now, there is an Equivalent definition to the above definition of Well-Ordered set which is "A total ordered structure is well-ordered if and only if it does not contain infinite descending chains; that is, a linearly ordered set $(S, ≤)$ is a well-ordered set if and only if there does not exist a sequence $a_0, a_1, a_2, . . .$ of elements of $S$ such that $a_0 > a_1 > a_2 > . . . .$ ..We will call it as Definition 2 of well order.
We can prove that Definition 1 and Definition 2 are Equivalent (Proof is given below, after the answer to the asked question).
Since now that we know both the definitions are equivalent. We can use the Definition 1 to check for Well-ordering.
Now consider the following Subset of the given set $A$ :
$S$ = $\left \{ a^nb\,\,|\,\,n \geq 0 \right \}$.. this Subset has No least element. Hence, The given Structure $(A, dictionary \,\,order)$ is NOT Well-Ordered Structure.
Proving that Definition 1 and Definition 2 are Equivalent :
Prove that a structure is well-ordered if and only if it does not contain infinite descending chains; that is, prove that a linearly ordered set (S, ≤) is a well-ordered set if and only if there does not exist a sequence a0, a1, a2, . . . of elements of S such that a0 > a1 > a2 > . . . .
(Only if Part) If $a_0 > a_1 > a_2$, . . . is an infinite descending sequence in $S$, then the set {${ a0, a1, a2, . . . }$} does not have a minimum element, so $S$ is not well-ordered.
(If part) Suppose that S is not well-ordered, and fix a nonempty $A ⊆ S$ that does not have a minimum element. Fix $a_0 ∈ A$. Since $a_0$ is not a minimum element of $A$, there exists $a_1 ∈ A$. such that $a_0 > a_1$. Since $a_1$ is not a minimum element of $A$, there exists $a_2 ∈ A$ such that $a_1 > a_2$. Continuing this construction inductively, we find an infinite descending chain in $A$.
