Equivalent definition for b-open sets. 
A set B is b-open if $B \subset Int(cl(B)) \cup Cl(Int(B))$

For semi open  S is semi open if $S\subset Cl(IntS)$ and equivalently if there exist an open set O such that $O\subset S \subset \bar{O}$.
Is there any equivalent condition for b-open set like  $O\subset B \subset \bar{O}$ and  $O^{\circ}\subset B \subset {O}$, or any other like this.
 A: Somewhat surprisingly, we can systematically rule out every possible characterization of the form
$$\text{A set }A\text{ is }b\text{-open iff there is a set }O\text{ such that }O'\subset A\subset O''$$
for $O',O''$ formed by any sequence of the interior and closure operations on $O$, because $Cl\circ Int(O)=Cl\circ Int\circ Cl\circ Int(O)$ and $Int\circ Cl(O)=Int\circ Cl\circ Int\circ Cl(O)$. Due to this equivalence, we only need to check the cases for $O',O''$ as they range over the possibilities
$$O,Int(O),Cl\circ Int(O), Int\circ Cl\circ Int(O), Cl(O), Int\circ Cl(O), Cl\circ Int\circ Cl(O).$$
First, note that $\mathbb{Q}$ is $b$-open because it is dense. Conversely, the Cantor set, $\mathcal C$, is not $b$-open because it is nonempty, closed, and contains no nonempty open set. Generally, we will use $\mathbb{Q}$ as an example of a $b$-open set that cannot satisfy inclusions that are too strong, and will use $\mathcal C$ as an example of a set that is not $b$-open but still satisfies inclusions that are too weak.
$$\text{Case 1: $O'$ assumes any possibility},\ O''=Int(O),Cl\circ Int(O),\text{ or }Int\circ Cl\circ Int(O)$$
Let $A=\mathbb{Q}$. $A$ contains $O'$ which contains $Int(O)$, and $A$ can contain no nonempty open set, so $Int(O)$ is empty, thus $O''$ is, so $A$ is a $b$-open set that cannot satisfy this inclusion.
$$\text{Case 2: } O'\text{ assumes on any possibility, }O''=O,Cl(O)$$
Let $A=\mathcal C=O$. Then $A$ is not $b$-open but satisfies this inclusion.
$$\text{Case 3: }O'=Int(O),Cl\circ Int(O), Int\circ Cl\circ Int(O),\ O''=Int\circ Cl(O), Cl\circ Int\circ Cl(O)$$
Let $A=\mathcal C$ and let $O$ be $\mathbb{Q}$. Then $A$ is not $b$-open but satisfies this inclusion.
$$\text{Case 4: }O'=O,\ O''=Int\circ Cl(O), Cl\circ Int\circ Cl(O)$$
If $O''=Int\circ Cl(O)$, then this characterizes pre-open sets, and if $O''=Cl\circ Int\circ Cl(O)$, then this characterizes $\beta$-open sets, which are distinct. As a counterexample to the second case, $A=\mathbb{Q}\cap [0,1)$ is not $b$-open but satisfies this inclusion for $O=A$. As a counterexample to the first case, this paper by Dimitrije Andrijević provides $A=[0,1]\cup ((1,2)\cap\mathbb{Q})$. $A$ is $b$-open, but $A\not\subset Int\circ Cl(A)$, so $A\not\subset Int\circ Cl(O)$ for any $O\subset A$.
$$\text{Case 5: }O'=Cl(O),Int\circ Cl(0),Cl\circ Int\circ Cl(O),\ O''=Int\circ Cl(O), Cl\circ Int\circ Cl(O)$$
Let $A=\mathbb{Q}$. Then $O'\subset A$ implies that $Int\circ Cl(O)$ is empty, so $O''$ is. Thus $A$ cannot satisfy this inclusion.
None of the cases above work, thus $b$-open sets cannot be characterized in this manner.

Of course, we have the following characterization if we are allowed to use two sets instead:
A set $A$ is $b$-open if there are sets $O,V$ such that $O$ is open and $O\cup V\subset A\subset Cl(O)\cup Int\circ Cl(V)$.

Finally, also see Proposition 2.1. of the paper mentioned above. It gives characterizations of $b$-open sets in terms of preclosure and preinterior operations, so that
$$\text{A set $A$ is $b$-open iff there is a set $O$ such that }O\subset A\subset pCl\circ pInt(O),$$
for $pInt(O)=O\cap Int\circ Cl(O)$ and $pCl(O)=O\cup Cl\circ Int(O)$. However, such characterizations likely only provide convenience in limited contexts as you still have to deal with taking unions and intersections of sets just as in the original definition.
