The equation I need to solve is
$\mathbf{R} = \mathbf{X}^T\mathbf{W}\mathbf{Y}$
where $\mathbf{R}$ is in $\mathbb{R}^{l \times m}$. $\mathbf{X}$, $\mathbf{W}$, and $\mathbf{Y}$ are in $\mathbb{R}^{a \times l}$, $\mathbb{R}^{a \times b}$ and $\mathbb{R}^{b \times m}$ respectively.
As some note I've seen, this equation could be transform to like (this may be incorrect)
$\mathbf{R} = \mathbf{W}^T(\mathbf{X} \otimes \mathbf{Y})$
using tensor product.
If $\mathbf{R}$, $\mathbf{X}$, and $\mathbf{Y}$ are given, $\mathbf{W}$ can be derived by general implemented regression software packages. this is the reason why transforming to the second equation.
Could I transform as above?
Here you have
$$\mathbb{R}^m \stackrel{Y}{\longrightarrow}\mathbb{R}^b \stackrel{W}{\longrightarrow}\mathbb{R}^a \stackrel{X^T}{\longrightarrow}\mathbb{R}^l $$
You can use the natural isomorphisms
$$\text{Hom}(\mathbb{R}^b, \mathbb{R}^a) \cong (\mathbb{R}^b)^{\star} \otimes \mathbb{R}^a \;\; \text{and} \;\; \text{Hom}(\mathbb{R}^m, \mathbb{R}^l) \cong (\mathbb{R}^m)^{\star} \otimes \mathbb{R}^l$$
and send the map $W \mapsto X^T W Y$ to $Y^T \otimes X^T $ and see $X^TWY=R$ is equivalent to $(Y^T \otimes X^T)W=R$ or $(Y \otimes X)^T W = R$ where $W \in (\mathbb{R}^b)^{\star} \otimes \mathbb{R}^a$ and $R \in (\mathbb{R}^m)^{\star} \otimes \mathbb{R}^l.$