How to solve this question? Show   that  if  $B$  is  a  basis for  a  topology  on  $X$ , then  the  topology  generated  by  $B$ is equal to the  intersection  of  all   topologies  on  $X$   that  contain   $B$.
 A: Perhaps, as joriki suggests, the question is to show that for a base $\mathcal{B}$ for a topology on $X$, the family $\tau_{\mathcal{B}}$ of arbitrary unions of elements of $\mathcal{B}$ -- including, I suppose, the empty union, which gives the empty set -- is the smallest topology containing $\mathcal{B}$.
If so, this is certainly true and can be shown as follows:
Step 1: Since any topology is closed under arbitrary unions, any topology $\tau$ containing $\mathcal{B}$ must contain $\tau_{\mathcal{B}}$, so it is enough to show that $\tau_{\mathcal{B}}$ is a topology.  
Recall that what we are assuming about $\mathcal{B}$ is that
(B1) $\bigcup_{B \in \mathcal{B}} B = X$ and
(B2) For all $B_1, B_2 \in \mathcal{B}$, if $x \in B_1 \cap B_2$, then there exists 
$B_3 \in \mathcal{B}$ such that $x \in B_3 \subset B_1 \cap B_2$.  
Step 2: Thus $\emptyset, X$ are unions of elements of $\mathcal{B}$: the former by taking the empty union, the latter by (B1).  
Step 3: Being the set of all unions of a certain family of sets, $\tau_{\mathcal{B}}$ is certainly closed under arbitrary unions.
Step 4: So the matter of it is to show that $\tau_{\mathcal{B}}$ is closed under finite intersections.  For this, it is enough to show that if $U_1,U_2 \in \tau_{\mathcal{B}}$, then so is $U_1 \cap U_2$.  To show this we need to use condition (B2), which notice has not yet been used.  This verification takes two or three lines.  I urge the OP to try it herself and tell us whether she succeeded and if not what she tried.
