I am under the impression that some of the answers given may be interpreting question (A) in a manner which does not seem --- at least to me --- entirely consistent with how it is stated. Interpreting question (A) quite strictly, it seems to me that there can be in fact two distinct differentiable structures on the total space of a topological vector bundle which both make it into a smooth vector bundle. In fact, for any non-empty manifold $B$ of dimension greater than zero, and any topological vector bundle $E$ over $B$ of dimension greater than zero, there exist uncountably many distinct smooth structures on $E$ for which $E$ becomes a smooth vector bundle over $B$. I will give a very simple (and detailed) example below. Also, I will work --- out of habit --- with real vector bundles, but the exact same construction with ${\mathbb R}$ replaced by ${\mathbb C}$ works equally well for complex line bundles.

Let $B$ be a manifold, and $E$ a *topological* vector bundle over $B$ with projection $\textrm{proj}:E\to B$ (as usual, we confuse the vector bundle with its total space). Assume the total space $E$ admits a differentiable structure which makes it into a smooth vector bundle over $B$. Denote this *smooth* vector bundle (and its total space seen as a smooth manifold) by $E^{(1)}$, to distinguish it from the *topological* vector bundle $E$. For any continuous map $f:B\to{\mathbb R}\setminus\{0\}$ we can construct the map $H_f:E\to E$ given by
$$ H_f(x)= f(\textrm{proj}(x))\cdot x $$
In other words, the map $H_f$ is the vector bundle map $E\to E$ which sits over the identity $\textrm{id}_B$ on $B$, and which is multiplication by $f(b)$ on the fibre over $b\in B$.

It is obvious that $H_f:E\to E$ is an isomorphism of *topological* vector bundles whose inverse is $H_{\frac 1 f}$ (since $f$ is never zero). But it does *not* give a map of *smooth* vector bundles $E^{(1)}\to E^{(1)}$ unless $f$ is itself smooth. Now transfer the smooth structure on $E^{(1)}$ via the homeomorphism $H_f$, and denote the new smooth manifold by $E^{(f)}$. More precisely, $E^{(f)}$ is the topological space $E$ equipped with the unique differentiable structure which makes $H_f:E_0\to E^{(f)}$ a diffeomorphism.

Recall that $\textrm{proj}:E^{(1)}\to B$ is a smooth vector bundle. Note also that $H_f$ is both a diffeomorphism $H_f:E^{(1)}\to E^{(f)}$ and an isomorphism of topological vector bundles $H_f:E\to E$. As a consequence, the *smooth* vector bundle structure on $\textrm{proj}:E^{(1)}\to B$ transfers across $H_f$ to a *smooth* vector bundle structure on $\textrm{proj}:E^{(f)}\to B$ whose underlying *topological* vector bundle is $E$. In other words, the topological vector bundle $E$ underlying $\textrm{proj}:E^{(f)}\to B$ is actually a smooth vector bundle when we consider the smooth structure $E^{(f)}$ on the total space.

To summarize, we have two smooth vector bundle structures, $E^{(1)}$ and $E^{(f)}$, on the topological vector bundle $E$ over $B$. [By the way, the notation is self-consistent: observe that for $f=1$, $E^{(f)}$ is just the original $E^{(1)}$.] To answer the question (A) affirmatively, and give an example at the same time, assume:

- the vector bundle $E$ has positive dimension;
- $B$ is non-empty and has positive dimension.

Then I claim:

**Lemma:** The identity function on $E$ is a diffeomorphism (or even just a smooth function) $E^{(f)}\to E^{(1)}$ if and only if $f$ is itself smooth. $\square$

On the one hand, it is easy to check that if $f$ is smooth then $H_f$ is a diffeomorphism $E^{(1)}\to E^{(1)}$ (both it and its inverse $H_{\frac 1 f}$ are smooth). Therefore, the smooth structure on $E^{(f)}$ is by definition the same as the smooth structure on $E^{(1)}$, i.e. the identity is a diffeomorphism $E^{(f)}\to E^{(1)}$. On the other hand, assume that the identity $\textrm{id}_E$ is a smooth function $E^{(f)} \to E^{(1)}$. I will prove that $f$ is itself smooth. Consider the composition of diffeomorphisms
$$ G : E^{(1)} \overset{H_f}{\longrightarrow} E^{(f)} \overset{\textrm{id}_E}{\longrightarrow} E^{(1)} $$
By definition of $H_f$:
$$ G(x) = f(\textrm{proj}(x))\cdot x $$
and it is fairly easy to use local trivializations for $E^{(1)}$ to conclude that $f$ is smooth. In fact, if $\varphi:U\times {\mathbb R}^n \to E$ is a (smooth) trivialization of $E^{(1)}$ over the open $U\subset B$, it follows that
$$ \varphi^{-1}\circ G\circ\varphi(u,v) = (u,f(u)\cdot v) $$
and since $E$ has positive dimension, the restriction of $f$ to $U$ is the last component of the smooth function
$$ u\longmapsto\varphi^{-1}\circ G\circ\varphi(u,(0,\ldots,0,1)) $$
Hence $f$ is smooth.

In conclusion, under condition 1 above, the smooth structure on $E^{(1)}$ is the same as the smooth structure on $E^{(f)}$ if and only if $f$ is smooth. Furthermore, under condition 2 above, we can then find a continuous map $f:B\to{\mathbb R}\setminus\{0\}$ which is not smooth. For such a choice of $f$, $E^{(1)}$ and $E^{(f)}$ are two distinct smooth vector bundle structures on $E$.

In fact, more can be said. If we start with $E^{(f)}$ in place of $E^{(1)}$ and apply the above construction, we can easily describe the result: for any continuous functions $f,g:B\to{\mathbb R}\setminus\{0\}$ it is easy to check that $H_g \circ H_f = H_{f\cdot g}$, which implies
$$ (E^{(f)})^{(g)} = E^{(f\cdot g)} $$
Applying the preceding lemma, we conclude that the smooth structure on $E^{(f)}$ coincides with the smooth structure on $E^{(g)}$ if and only if $\frac f g$ is smooth. Consequently, the construction $f\mapsto E^{(f)}$ gives a set of smooth vector bundle structures on $E$ which is in bijection with the quotient of abelian groups
$$ C^0(B,{\mathbb R}\setminus\{0\})/C^\infty(B,{\mathbb R}\setminus\{0\}) $$
where the multiplication in each of the groups is given by multiplying functions. It is fairly easy to see that this quotient is uncountable under condition 2 above: by giving for each $b\in B$ a continuous function $f_b:B\to{\mathbb R}\setminus\{0\}$ which is smooth everywhere except at the point $b$, we determine an injection of $B$ into the above quotient of abelian groups.

### Essential uniqueness of the smooth vector bundle structure on $E$

In light of the above, what can be said regarding uniqueness of the smooth vector bundle structure on $E$? Well, as some of the other answers have indicated, one can use approximation of continuous functions by smooth functions to prove the following result from its topological counterpart (i.e. the usual topological classification of vector bundles).

**Theorem:** Let $B$ be a smooth manifold of dimension $n$. Consider the function
$$ \theta:[B,\textrm{Gr}(k,{\mathbb R}^{k+l})]^{\textrm{smooth}}\longrightarrow \textrm{Vec}^{\textrm{smooth}}_B $$
(where the domain is the set of smooth homotopy classes of smooth functions from $B$ into the Grassmannian of $k$-dimensional linear subspaces of ${\mathbb R}^{k+l}$, and the target is the set of isomorphism classes of *smooth* $k$-dimensional vector bundles over $B$) defined by
$$ \theta([f])=f^\ast(\gamma_{k,k+l}) $$
where $\gamma_{k,k+l}$ is the tautological smooth vector bundle over the Grassmannian. Then $\theta$ is a bijection if $l\geq n+2$. $\square$

By using this theorem, its topological counterpart (replace smooth by continuous/topological), and the approximation of continuous functions by smooth functions, one can see that the forgetful map from the set of isomorphism classes of smooth vector bundles over $B$ to the set of isomorphism classes of topological vector bundles over $B$ is a bijection.

We do not really need the above theorem to see that isomorphism classes of smooth vector bundles inject into the isomorphism classes of topological vector bundles (although it is a convenient way to prove surjectivity). Using only the approximation of continuous function by smooth functions and smooth partitions of unity, one can approximate any topological isomorphism between smooth vector bundles by a smooth isomorphism. Moreover, given isomorphisms $\varphi,\psi:E\to E'$ of smooth vector bundles over $B$, any homotopy through topological isomorphisms between $\varphi$ and $\psi$ can be approximated by a homotopy through smooth isomorphisms between $\varphi$ and $\psi$. In particular, given two smooth vector bundle structures $E_1$ and $E_2$ on a topological vector bundle $E$ over $B$, there exists a *unique* homotopy class of smooth vector bundle isomorphisms $E_1\to E_2$ which, as an isomorphism of topological vector bundles, is homotopic to the identity $\textrm{id}_E:E\to E$.

Continuing in the same manner, it is not too hard to conclude that the homotopy fibres of the map
$$ \textrm{Iso}^{\textrm{smooth}}_B(E_1,E_2) \longrightarrow \textrm{Iso}^{\textrm{top}}_B(E,E) $$
(between the spaces of isomorphisms of smooth/topological vector bundles over $B$) are weakly contractible. Consequently, the map is a weak equivalence. That appears to be the strongest result we can state, to the best of my current knowledge, and especially in light of my examples above.