Here's my two cents although it's rather sketchy.
For any CW complex $X$, $H^3(X;mathbb{Z})=[X,K(mathbb{Z},3)]$, where $K(mathbb{Z},3)$ comes equipped with a fibration $mathbb{CP}^inftyto Pto K(mathbb{Z},3)$. The total space $P$ is contractible. Now suppose $X$ is a compact manifold of dimension $n$ which is $2$-connected and $H^3(X;mathbb{Z})=mathbb{Z}$. Then choosing a generator of $H^3(X;mathbb{Z})$ corresponds to a (homotopy class of) map $f:Xto K(mathbb{Z},3)$. The pullback bundle $f^ast Pto X$ has the property that $H^3(f^ast P;mathbb{Z})=0$.
Since we need a finite dimensional manifold which $f^ast P$ isn't, let $E$ denote the $(n+5)$-skeleta of $f^ast P$. It is compact and locally looks like $Xtimesmathbb{CP}^2$. I think(?) that $pi:Eto X$ is a fibre bundle. Since $pi_3$ is unchanged for $4$-skeleta or higher, it follows that $0=pi_3(E)=pi_3(f^ast P)$, whence $H^3(E;mathbb{Z})=0$.
Feel free to tweak the answer if need be.
Edit As pointed out by algori and Igor, the second paragraph doesn't give you a fibre bundle.
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