The answer is **YES** (assuming $B$ is a **smooth manifold**). A proof can be found on Walschap's book "[Metric Structures in Differential geometry][1]", p. 77, Lemma 7.1.

For the OP's convenience, here's a sketch of the proof. Choose an open cover of $B$ such that your vector bundle is trivial over each element. From general results in topology, this (and in fact any) cover of an $n$-dim manifold $B$ admits a refinement $\lbrace{V_\alpha\rbrace}_{\alpha\in A}$

such that any point in $B$ belong to at most $n+1$ $V_\alpha$'s. Let $\{\phi_\alpha\}$ be a partition of unity subordinate to this cover and denote by $A_i$ the collection of subsets of $A$ with $i+1$ elements. Given $a=\{\alpha_0,\dots,\alpha_i\}$$\in A_i$, denote by $W_a$ the set consisting of those $b\in B$ such that $\phi_\alpha(b)<\phi_{\alpha_0}(b),\dots,\phi_{\alpha_i}(b)$ for all $\alpha\neq\alpha_0,\dots,\alpha_i$. Then the collection of open subsets $U_i:=\cup_{a\in A_i} W_a$ covers $B$ and is such that your bundle restricted to each $U_i$ is trivial.


  [1]: http://www.amazon.com/Metric-Structures-Differential-Geometry-Walschap/dp/038720430X