a simple question about topological space X is a topological space, $ A\subseteq B\subseteq X $, if $A$ is a nowhere dense subset of $B$ , then $A$ is a nowhere dense subset of the whole space $X$?

Is this right? I thind it's right, but I can't prove it. Maybe we need to add some contions: e.g. $B$ is a closed subset of $X$, and/or $A$ is (also) a closed subset of $X$.
If $A$ is nowhere dense in $B$, then $\operatorname{cl}_BA$ has empty interior in $B$, and therefore $B\setminus\operatorname{cl}_BA$ is a dense open set in $B$. There is then a $U$ open in $X$ such that $U\cap B=B\setminus\operatorname{cl}_BA$. Let 

$$V=U\cup(X\setminus\operatorname{cl}_XB)\;;$$

clearly $V$ is open in $X$, and you should have no trouble showing that $V$ is dense in $X$. Finally, $V\cap A=\varnothing$, so $A$ must be nowhere dense in $X$.
If $\overline{B}$ has empty interior, then since $\overline{A} \subseteq \overline{B}$, $\overline{A}$ must have empty interior also (otherwise contradicting the nowhere-denseness of $B$).