Let $T=\set{\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3,\,\vect{w}_4}$. The vector $\colvector{2\\-1\\0\\1}$
is a solution to the homogeneous system with the matrix $B$ as the coefficient matrix (check this!). By Theorem SLSLC it provides the scalars for a linear combination of the columns of $B$ (the vectors in $T$) that equals the zero vector, a relation of linear dependence on $T$,
\begin{equation*}
2\vect{w}_1+(-1)\vect{w}_2+(1)\vect{w}_4=\zerovector
\end{equation*}
We can rearrange this equation by solving for $\vect{w}_4$,
\begin{equation*}
\vect{w}_4=(-2)\vect{w}_1+\vect{w}_2
\end{equation*}
This equation tells us that the vector $\vect{w}_4$ is superfluous in the span construction that creates $W$. So $W=\spn{\set{\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3}}$. The requested set is $R=\set{\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3}$.