= Solution
\b[False even when both base and fibre are closed and oriented.] Take the <Hopf fibration> $h:S^3\to S^2$, and define the smooth <fiber bundle> $p:S^3\times S^1\to S^2$ by $p(x,\theta)=h(x)$. Its local trivializations are the Hopf trivializations times $S^1$, and its fibre is $S^1\times S^1=T^2$.
The <Künneth theorem> for <cohomology> over a field says $H^k(A\times B;\mathbb R)=\bigoplus_{i+j=k}H^i(A;\mathbb R)\otimes H^j(B;\mathbb R)$ for these compact manifolds. Here it gives $H^2(S^3\times S^1;\mathbb R)=0$. If a <symplectic form> $\omega$ existed on this closed four-manifold, it would be exact, $\omega=d\eta$. Since $d\omega=0$,
$$
\int_{S^3\times S^1}\omega\wedge\omega
=\int_{S^3\times S^1}d(\eta\wedge\omega)=0
$$
by <Stokes theorem>. But $\omega^2$ is a positive volume form in the <symplectic orientation>, so its integral is positive. This contradiction proves the claim fails. The mere bundle structure does not ensure that a fibrewise area class extends to the total space.
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