Magnetic Field calculated from Magnetic vector potential Reference : $^{\prime\prime}$Physics of Atoms and Molecules$^{\prime\prime}$ by B.H.Bransden and C.J.Joachain, 1983 Edition.
In $\boldsymbol{\S}$4.1 The electromagnetic field and its interaction with charged particles we read :

\begin{align}
\mathscr{E}\!\!\!\!\mathscr{E} \left(\mathbf{r},t\right) & \boldsymbol{=}\boldsymbol{-}\boldsymbol{\nabla}\phi\left(\mathbf{r},t\right)\boldsymbol{-}\dfrac{\partial}{\partial t}\mathbf{A}\left(\mathbf{r},t\right)
\tag{4.1}\label{4.1}\\ 
\mathscr{B}\!\!\!\!\!\mathscr{B} \left(\mathbf{r},t\right) & \boldsymbol{=}\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}\left(\mathbf{r},t\right)
\tag{4.2}\label{4.2} 
\end{align}
\begin{align}
\boldsymbol{\nabla\cdot}\mathbf{A} & \boldsymbol{=}0
\tag{4.3}\label{4.3}\\ 
\nabla^2\mathbf{A} \boldsymbol{-}\dfrac{1}{c^2}\dfrac{\partial^2\mathbf{A}}{\partial t^2} & \boldsymbol{=0}
\tag{4.4}\label{4.4} 
\end{align}
$\cdots\cdots\cdots\cdots$the most general solution of Maxwell's equations for a radiation field can always be expressed in terms of potentials such that $\boldsymbol{\nabla\cdot}\mathbf{A}\boldsymbol{=}0$ and $\phi\boldsymbol{=}0$.


A monochromatic plane wave solution of equations \eqref{4.3}-\eqref{4.4}
corresponding to the angular frequency $\omega$ (i.e. to the frequency $\nu=\omega/2\pi$) is one that represents a real vector potential $\mathbf{A}$ as
\begin{align}
\mathbf{A}\left(\omega;\mathbf{r},t\right) & \boldsymbol{=}2\mathbf{A}_{0}\left(\omega\right)\cos\left(\mathbf{k}\boldsymbol{\cdot}\mathbf{r}\boldsymbol{-}\omega t\boldsymbol{+}\delta_{\omega}\right)
\nonumber\\ 
&\boldsymbol{=}\hphantom{2}\mathbf{A}_{0}\left(\omega\right)\left[\exp\left[i\left(\mathbf{k}\boldsymbol{\cdot}\mathbf{r}\boldsymbol{-}\omega t\boldsymbol{+}\delta_{\omega}\right)\right]\boldsymbol{+}\text{c.c.}\right]
\tag{4.5}\label{4.5} 
\end{align}
Here $\mathbf{A}_{0}$ is a vector which, as we shall see shortly, describes both the intensity and the polarisation of the radiation, $\mathbf{k}$ is the propagation vector, $\delta_{\omega}$ is a real phase and c.c. denotes the complex conjugate.We note that \eqref{4.3} is satisfied if
\begin{equation}
\mathbf{k}\boldsymbol{\cdot}\mathbf{A}_{0}\left(\omega\right)\boldsymbol{=}0
\tag{4.6}\label{4.6}   
\end{equation}
so that $\mathbf{A}_{0}$ is perpendicular to $\mathbf{k}$ and the wave is transverse. Equation \eqref{4.4} is satisfied provided that $\omega\boldsymbol{=}k\,c$, where $k$ is the magnitude of the propagation vector $\mathbf{k}$.


$\cdots\cdots\cdots\cdots$


The electric and magnetic fields associated with the vector potential \eqref{4.5} are given respectively from \eqref{4.1} and \eqref{4.2} by
\begin{align}
\mathscr{E}\!\!\!\!\mathscr{E}  & \boldsymbol{=}\boldsymbol{-}2\omega A_{0}\left(\omega\right)\boldsymbol{\hat{\varepsilon}}\sin\left(\mathbf{k}\boldsymbol{\cdot}\mathbf{r}\boldsymbol{-}\omega t\boldsymbol{+}\delta_{\omega}\right) 
\nonumber\\  
\mathscr{B}\!\!\!\!\!\mathscr{B} & \boldsymbol{=}\boldsymbol{-}2 A_{0}\left(\omega\right)\left(\mathbf{k}\boldsymbol{\times}\boldsymbol{\hat{\varepsilon}}\right)\sin\left(\mathbf{k}\boldsymbol{\cdot}\mathbf{r}\boldsymbol{-}\omega t\boldsymbol{+}\delta_{\omega}\right)
\tag{4.7}\label{4.7} 
\end{align}
where we have written $\mathbf{A}_{0}\left(\omega\right)\boldsymbol{=}A_{0}\left(\omega\right)\boldsymbol{\hat{\varepsilon}}$. The direction of the electric field $\mathscr{E}\!\!\!\!\mathscr{E}$ is along the real unit vector $\boldsymbol{\hat{\varepsilon}}$, which specifies the polarisation of the radiation, and is called the polarisation vector. From \eqref{4.6} we note that $\boldsymbol{\hat{\varepsilon}}$ must lie in a plane perpendicular to the propagation vector $\mathbf{k}$ and can therefore be specified by
giving its components along two linearly independent vectors lying in this plane.
We also see from \eqref{4.7} that both $\mathscr{E}\!\!\!\!\mathscr{E}$ and $\mathscr{B}\!\!\!\!\!\mathscr{B}$ are perpendicular to the direction of
propagation $\mathbf{k}$, and to each other.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
Question : How the expression for $\mathscr{B}\!\!\!\!\!\mathscr{B}$ in \eqref{4.7} is derived  from \eqref{4.2} and \eqref{4.5} ?
 A: Hint :
If $\psi\left(\mathbf{r}\right), \boldsymbol{\alpha}\left(\mathbf{r}\right)$ scalar and vector functions of $\mathbf{r}$ respectively then
\begin{equation}
\boldsymbol{\nabla}\boldsymbol{\times}\left(\psi\,\boldsymbol{\alpha}\right)\boldsymbol{=}\boldsymbol{\nabla}\psi\boldsymbol{\times}\boldsymbol{\alpha}\boldsymbol{+}\psi\boldsymbol{\nabla}\boldsymbol{\times}\boldsymbol{\alpha}
\tag{Fr-01}\label{Fr-01}   
\end{equation}
So if $\boldsymbol{\alpha}$ is a constant vector independent of $\mathbf{r}$ then
\begin{equation}
\boldsymbol{\nabla}\boldsymbol{\times}\left(\psi\,\boldsymbol{\alpha}\right)\boldsymbol{=}\boldsymbol{\nabla}\psi\boldsymbol{\times}\boldsymbol{\alpha}\,,\qquad\boldsymbol{\alpha}\boldsymbol{=}\textbf{constant} 
\tag{Fr-02}\label{Fr-02}   
\end{equation}
Inserting in \eqref{Fr-02} $\boldsymbol{\alpha}\boldsymbol{=}2\mathbf{A}_{0}\left(\omega\right)\boldsymbol{=}2A_{0}\left(\omega\right)\boldsymbol{\hat{\varepsilon}}\boldsymbol{=}\textbf{constant}$ and  $\psi\left(\mathbf{r}\right)\boldsymbol{=}\cos\left(\mathbf{k}\boldsymbol{\cdot}\mathbf{r}\boldsymbol{-}\omega t\boldsymbol{+}\delta_{\omega}\right)$ the only thing you must try to do is to find the grad of a scalar function (more simple than the curl of a vector function).
A: The result given by the OP is off by a sign (v1). It should read
$$\vec{B}=-2{A_0}(\hat{k}\times\hat{\epsilon})\sin({\textbf{k.r}-\omega t+\delta_{\omega}}).$$
Let me do one component. If $\hat{\epsilon}=(\epsilon_x,\epsilon_y,\epsilon_z)$, then
\begin{align}
(\nabla\times\vec{A})_x&=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\\
&=2A_0\epsilon_z\left[-\sin(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_\omega)\right]k_y-2A_0\epsilon_y\left[-\sin(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_\omega)\right]k_z\\
&=2A_0\sin(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_\omega)[\epsilon_yk_z-\epsilon_zk_y]\\
&=2A_0\sin(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_\omega)(\hat{\epsilon}\times\hat{k})_x\\
&=-2A_0\sin(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_\omega)(\hat{k}\times\hat{\epsilon})_x.
\end{align}
To see this in a more general way, remember that the $i$-th component of a curl is
$$(\nabla\times\vec{A})_i=\sum_{j,k}\varepsilon_{ijk}\frac{\partial A_k}{\partial x_j},$$
and that the cross product can be written as
$$(\vec{k}\times\vec{\epsilon})_i=\sum_{j,k}\varepsilon_{ijk}k_j\epsilon_k.$$
By applying the chain rule we get
$$\frac{\partial A_k}{\partial x_j}=-2A_0\epsilon_k\sin(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_\omega)k_j.$$
It is immediate to get to the general result by making use of the last three equations.
