Let $A=[v_1\;v_2\;v_3\;v_4]$ be the required matrix. You need that $$ A\begin{bmatrix}1\\\2\\\0\\\0\end{bmatrix}=v_1+2v_2=0 $$ so $v_1=-2v_2$; also $$ A\begin{bmatrix}0\\\1\\\2\\\0\end{bmatrix}=v_2+2v_3=0 $$ so $v_2=-2v_3$ and $v_1=4v_3$.
You see that $v_4$ can be anything; now just take $$ v_3=\begin{bmatrix}1\\\2\\\0\\\0\end{bmatrix} \quad v_4=\begin{bmatrix}0\\\1\\\2\\\0\end{bmatrix} $$ Then the matrix $$ A=[v_1\;v_2\;v_3\;v_4]= \begin{bmatrix} 4 & -2 & 1 & 0 \\\ 8 & -4 & 2 & 1 \\\ 0 & 0 & 0 & 2 \\\ 0 & 0 & 0 & 0 \end{bmatrix} $$ will have the required column space and null space: indeed, the column space is generated by $v_3$ and $v_4$, so it has dimension $2$. Since the null space contains the required one, it will be equal to it by the rank-nullity theorem.