Find a singular value decomposition for $$X = \begin{bmatrix} -2 & 2 \\ -1 & 1 \\ 2 & -2 \end{bmatrix}$$ So far I have found $$X^tX = \begin{bmatrix} 9 & -9 \\ -9 & 9 \\ \end{bmatrix}$$ with eigenvalues $λ = 18$ and $0$. I was then able to find $$u_1 = \begin{bmatrix} \tfrac23 \\ \tfrac13 \\ -\tfrac23 \end{bmatrix}$$ As the second eigenvalue is $0$, I then could not find $u_2$ and $u_3$ in order to work out $U\Sigma V$ and thus find the final answer.
How can I find $u_2$ and $u_3$ to complete the matrix $U$?
You don't need $u_3$, you just need $u_2$, and just like you found $u_1$ as the vector that satisfies the equation $X^TXu_1 = \lambda_1 u_1$, you can find $u_2$ as the vector that satisfies the equation $X^TXu_2=\lambda_2 u_2$, where, of course, $\lambda_1$ and $\lambda_2$ are the two singular values you found.