Is the category of class of all sets a concrete category? Since the class of all sets is not a set but is  a class, is the category of  class of all sets a concrete category?
(the only object is the class of all sets)
 A: Your comments suggest that you have a worrying misconception about category theory: specifically, your attempt to say that the single object of your category 'is' something. The entire point of category theory is that we can study objects by only looking at the morphisms between them, without looking at the internal structure of the objects. Looking at it like that, you can see why we are worried by your insistence that your object 'is' $\underline{Set}$: once you use it as an object, it's just that object, and nothing else.
Objects in category theory derive their meaning from the morphisms that start or end there; vice-versa, the only reason we describe objects as 'being' certain things is in order to help us understand the morphisms. A statement like, "the objects of this category are groups," is inevitably followed up with, "the morphisms of this category are group homomorphisms." Once you define a category as having morphisms which are group homomorphisms acting in the way you expect, the objects look like groups whether you intended them to or not.
So, when you say that your category has $\underline{Set}$ as the only object, that means little. All we know is that your category has only one object; that object derives its meaning from the morphisms, and if we choose those morphisms without referring to $\underline{Set}$ (e.g. Neptun's answer, where the only morphism is the identity) the object stops looking like $\underline{Set}$ at all, because the morphisms don't require it.
The only way to give the object the 'meaning' of being $\underline{Set}$ is to choose a class of morphisms that give it that meaning, which is why everyone has been asking about the morphisms that you want. Two possible examples are:


*

*Your category has as morphisms the endofunctors from the category of sets to itself. This is not concrete for mere size reasons (this category is not locally small).

*Your category has as morphisms equivalence classes of endofunctors under natural isomorphism. This is not concrete for the same mere reason (it's still not locally small), but also for a slightly deeper reason that the category of (small) categories with equivalence classes of functors is known to not be concrete (which is an obstacle to trying to interpret this as concrete).


Either way, we cannot answer your question until we and you understand the morphisms you want to use.
A: A concrete category is a category with a faithful functor to Set, right? And faithful means that it is injective on the hom-sets for every pair of objects.
Seeing as the category with Set as the only object and the identity as the only arrow only has one hom-set, with one element, we can just map it to a singleton set $*$, and map the identity $1_\textbf{Set}$ to the identity $1_*$. And so the answer is yes!
A: I'm assuming the situation is as follows: We have a category $\mathcal{C}$ with one object $*$ and for morphisms we take the collection of functors $Set \to Set$. For this to be a concrete category we would need a functor $\mathcal{C} \to Set$ such that the corresponding map on homs is injective. But the collection of morphisms $* \to *$ is not a set! To see this, consider the functor $Set \to Set$ that sends every object to a given set $S$, and morphisms to the identity $S \to S$. This gives us an injection $V \to \operatorname{Hom}(*,*)$, where $V$ is the universe of all sets. Since $V$ is not a set, we're done. 
A similar argument works if we take $\operatorname{Hom}(*,*)$ to be the collection of functors $Ord \to Ord$, from the category of ordinals to itself (I mention this since it was asked in the comments).
