Find a general control and then show that this could have been achieved at x2
First of all we have forall $n\in N:$ $u_{n}\in R^{2}$. So let $$u_{0}=(u_{0,1},u_{0,2})^{t}$$ be given as well as $x_{0}=0$. Then we get $$ x_{1}=Ax_{0}+Bu_{0}=0+(0,u_{0,2},u_{0,1})^{t}$$ Again plugging this into the recursion we obtain $$ x_{2}=Ax_{1}+Bu_{1}=A(0,u_{0,2},u_{0,1})^{t}+B(u_{1,1},u_{1,2}).$$ This has by assumption to be $x_{3}$ so you get $4$ variables to determine. The system is underdetermined, but this should be no problem since you only have to solve the linear equation $$x_{3}=A(0,u_{0,2},u_{0,1})^{t}+B(u_{1,1},u_{1,2})$$ with respect to $u_{0,1},u_{0,2},u_{1,1},u_{1,2}$.
An update to your problem. The Kalman matrix $C$ is correct. Now we want to find the vector $u:=(u_{0,1},u_{0,2},u_{1,1},u_{1,2},u_{2,1},u_{2,2})^{t}\in R^{6}$ such that $$Cu=x_{3}.$$ We can write this in matrix vector representation as $$ \begin{pmatrix} 0&0&2&2&6&6&|2\\ 0&1&-1&0&-3&-2&|1\\ 1&0&1&0&1&0&|2 \end{pmatrix} $$ By changing rows and dividing by 2 $$ \begin{pmatrix} 1&0&1&0&1&0&|2\\ 0&1&-1&0&-3&-2&|1\\ 0&0&1&1&3&3&|1\\ \end{pmatrix} $$ This system is in echelon form. So we first determine one specifical solution, for instance $$u^{*}=\begin{pmatrix} 2\\1\\0\\1\\0\\0\end{pmatrix} $$. Next we need the KERNEL of the matrix, that is the elements $$\operatorname{ker}(C):=\{x\in R^{6}: Cx=0\}.$$ In matrix vector representation this is $$ \begin{pmatrix} 1&0&1&0&1&0&|0\\ 0&1&-1&0&-3&-2&|0\\ 0&0&1&1&3&3&|0\\ \end{pmatrix} $$ Since this matrix has $\operatorname{Rank}(C)=3$, we have to find $3$ vectors, linear independent and satisfying the kernel condition: The vectors $$v_{1}=\begin{pmatrix} 2\\0\\-3\\0\\1\\0 \end{pmatrix} \quad v_{2}=\begin{pmatrix} 0\\2\\0\\-3\\0\\1\end{pmatrix} \quad v_{3}=\begin{pmatrix} 1\\-1\\1\\-1\\-2\\2\end{pmatrix} $$ do the job, so they generate the kernel of $C$. Altogether we have as solution $$ u\in\{u^{*}+\lambda v_{1}+\mu v_{2}+\sigma v_{3}, \lambda,\mu,\sigma\in R\}. $$ So if you pick a control vector $u$ in this set, you control the state to $x_{3}$.