TITLE: Examples and properties of spaces with only trivial vector bundles
QUESTION [7 upvotes]: Let $B$ be a paracompact space with the property that any (topological) vector bundle $E \to B$ is trivial. What are some non-trivial examples of such spaces, and are there any interesting properties that characterize them?
For simple known examples we of course have contractible spaces, as well as the 3-sphere $S^3$. This one follows from the fact that its rank $n$ vector bundles are classified by $\pi_3 (BO(n)) = \pi_2 (O(n)) = 0$. I'm primarily interested in the case where $B$ is a closed manifold. Do we know any other such examples?
There is this nice answer to a MSE question which talks about using the Whitehead tower of the appropriate classifying space to determine whether a bundle is trivial or not. This seems like a nice tool (of which I am not familiar with) to approaching this problem. As a secondary question, could I ask for some insight/references to this approach?
EDIT Now that we know from the answers all the examples for closed $3$-manifolds, I guess I can now update the question to the case of higher odd dimensions. Does there exist a higher dimensional example?

REPLY [4 votes]: Here is another obstruction.

Suppose $M^n$ is a closed simply connected manifold which admits only trivial vector bundles.  Then $M$ cannot be a $\mathbb{Z}/2\mathbb{Z}$-homology sphere, unless $n=3$.

I'm not sure if the hypothesis that $M$ is simply connected is necessary, but it's certainly necessary in the proof.
Together with the end of Michael Albanese's answer, this implies that every simply connected $5$-manifold admits a non-trivial vector bundle.
Proof Sketch:  By the work in the other answers, we already know that if $M$ admits only trivial vector bundles, then $M$ is odd dimensional, orientable, and it has the rational homology of a sphere.  Thus, we may assume $n\geq 5$.
Now, assume for a contradiction that all the torsion in $H^\ast(M)$ is of odd order.
Let $k$ denote the least common multiple of the orders of the torsion and note that $k$ is odd.
Because $M$ is orientable, it has a degree $1$ map $f:M\rightarrow S^n$.  We will construct a non-trivial vector bundle on $M$ by pulling back a non-trivial bundle $E$ over $S^n$ along $f$.  When $n\neq 7$, we may use the tangent bundle $E=TS^n$.  When $n=7$, we let $E$ denote the rank $3$ vector bundle corresponding to a generator of $\pi_7(BO(3))\cong \pi_6(O(3)) = \pi_6(S^3) = \mathbb{Z}/12\mathbb{Z}$.  Let $\phi:S^n\rightarrow BO(s)$ denote the classifying map of the bundle $E$ (where $s = n$ if $n\neq 7$, and $s= 3$ when $n = 7$.)
We claim that $f^\ast E$ is a non-trivial vector bundle over $M$.  To see this, note that since $M$ is simply connected and all torsion is annihilated by $k$, there is a map $g:S^n\rightarrow E$ of degree $k^r$ for some integer $r\geq 1$.  (See the answer here for a proof.)
It is enough to show that $g^\ast(f^\ast E)$ is a non-trivial vector bundle over $S^n$.  Of course, it is enough to show that the map $\phi\circ g\circ f:S^n\rightarrow BO(s)$ is homotopically non-trivial. We'll show this by showing the induced map on $\pi_n$ is non-trivial.
So, consider the induced map on $\pi_n$.  For $n\geq 5$ odd (except $n=7$), Kervaire has shown that $\pi_n(BO(n))\cong \pi_{n-1}(O(n))$ is either $\mathbb{Z}/2\mathbb{Z}$ or $(\mathbb{Z}/2\mathbb{Z})^2$.  When $n\neq 7$, the fact that $E$ is non-trivial implies that $\phi_\ast$  is non-trivial, so, for $n\neq 7$, the kernel of the map $\phi_\ast:\pi_n(S^n)\rightarrow \pi_n(BO(n))$ is the even integers.  On the other hand, by our choice of $E$ when $n=7$, we have $\ker \phi_\ast = 12\mathbb{Z}$.  In either case, the image of $(g\circ f)_\ast:\pi_n(S^n)\rightarrow \pi_n(S^n)$ is multiplication by the odd number $k^r$, so not contained in $\ker \phi_\ast$.  The completes the sketch. $\square$.