Solution: Note that the complement of $A$ in $X$, that is $X\setminus A$ is open as its complement contains only finitely many points. Therefore, $A$ must be closed and hence, $\bar{A} = A$.

We claim that $\overset{\kern 0.1cm \circ}{A} = \emptyset$ and $\partial A = \mathbb{R} ^2$. To show that, let $(x,y) \in \overset{\kern 0.1cm \circ}{A} $. If $(x,y) \in \overset{\kern 0.1cm \circ}{A} $, then there exists $r>0$ such that the open ball $B((x,y),r) \subseteq A$. But this can not be possible as $\mathbb{R} \setminus \mathbb{Q} $ is dense in $\mathbb{R} $, that is, the ball $B((x,y),r)$ must contain a point $(a,b)$ where $a \in \mathbb{R} \setminus \mathbb{Q} $.

To show that the boundary is all of $\mathbb{R} ^2$, let $(x,y) \in \mathbb{R} ^2$. We need to show that an open ball $B((x,y),r)$ must intersect both $A$ and its compliment. It intersects the set $A$ because $\mathbb{Q} $ is dense in $\mathbb{R} $ and it intersects the compliment as $\mathbb{R} \setminus \mathbb{Q} $ is dense in $\mathbb{R} $.