Given a magnetic field how to find its vector potential? Is there an "inverse" curl operator? For a certain (divergenceless) $\vec{B}$ find $\vec{A} $ such that $\vec{B}= \nabla \times \vec{A} $.
Is there a general procedure to "invert"  $\vec{B}= \nabla \times \vec{A} $? An inverse curl?
(I was thinking of taking the curl of the previous equation:
$$ \nabla \times \vec{B}= \nabla \times \nabla \times \vec{A} = 0.  $$
Then using the triple cross product identity $ \nabla \times \nabla \times \vec{V} = \nabla (\nabla \cdot V) - \nabla^2 V$ but that does not quite simplify things... I was hoping to get some sort of Laplace equation for $\vec{A}$ involving terms of $\vec{B}$.)
 A: No, there is no inverse curl operator. In that way it is just like the ordinary derivative: $f(x)$ cannot uniquely be determined by integrating $f'(x)$. In general the relation $\vec{B} = \nabla \times \vec{A}$ defines a set of differential equations, which don't uniquely determine $\vec{A}$. For example you can add any curl-less vector field to $\vec{A}$ without changing $\vec{B}$. Boundary conditions can reduce this redundancy for example.
Usually a "gauge" is chosen for $\vec{A}$ which reduces the redundancy in $\vec{A}$. But not all gauges uniquely determine $\vec{A}$, either: Take for example one of the most common gauges, the Coulomb Gauge, whose condition is:
$$\nabla \cdot \vec{A} = 0$$
Notice that you can still add a constant vector to $\vec{A}$ and it will fulfill both its divergence and curl conditions.
Edit: I think the answer by Nullius in Verba is great. Once you have your boundary conditions and/or gauge, this can be used to find the corresponding vector potential. Note that also in many physics problems you can just guess the solutions to these differential equations - not as intimidating as it sounds.
A: You could try using the Helmholtz decomposition.
If $F$ is a twice-differentiable vector field on a bounded volume $V$ with boundary $S$, then it can be decomposed into divergence-free and curl free components.
$$F=-\nabla\Phi+\nabla\times \mathbf{A}$$
where
$$\Phi(\mathbf{r})=\frac{1}{4\pi}\int_V\frac{\nabla'\cdot F(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}dV'-\frac{1}{4\pi}\oint_S\mathbf{\hat{n}}'\cdot\frac{F(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}dS'$$
$$\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\int_V\frac{\nabla'\times F(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}dV'-\frac{1}{4\pi}\oint_S\mathbf{\hat{n}}'\times\frac{F(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}dS'$$
$\mathbf{\hat{n}}'$ is the unit outward normal and $\nabla'$ is the gradient with respect to $\mathbf{r}'$ rather than $\mathbf{r}$.
The curl on its own does not have a uniquely-defined inverse. However, the curl and divergence can be combined into a single operator that does have a unique inverse up to boundary conditions.
If the fields are assumed to approach zero at infinity, the boundary integral in each of the above expressions becomes zero. If the field is required to be solenoidal, then the divergence in the first expression will be zero.
A: You were very close in taking the curl and looking for a Laplace (actually Poisson) equation for $\mathbf{A}$.
You are allowed to assume that $\boldsymbol{\nabla} \cdot \mathbf{A} = 0$ (Coulomb gauge, as mentioned in doublefelix's answer). Then, using your triple cross product identity, you get $\boldsymbol{\nabla} \times \mathbf{B} = -\nabla^2\mathbf{A}$.
You can solve this for $\mathbf{A}$ using the Green's function $-1/4\pi r$ for the Laplacian, just as in electrostatics. This will give the Helmholtz result for $\mathbf{A}$ as in Nullius's answer (the $\Phi$ part of the Helmholtz decomposition will be zero here because $\boldsymbol{\nabla} \cdot \mathbf{B} = 0$).
