Isospin. It looks like a homework problem, with a summary answer in Okun's book, p 63. You need quark diagrams like a hole in the head.
This strangeness-changing decay violates isospin by 1/2, so assuming the $\Delta T=1/2$ piece of the hamiltonian dominates, and, since Λ is an isosinglet (so irrelevant to the isospin amps), you just consider the addition of a spurion s of isospin $|1/2,-1/2\rangle$ added to the cascade isodoublet $|1/2,\pm 1/2\rangle$, to yield pion states $|1,0\rangle$ and $|1,-1\rangle$ respectively.
The ratio of the respective decay amplitudes, then, is the simplest Clebsch ever, wich is exactly why your PDG booklet you no doubt are staring at has a Clebsch table: the most useful page of it! $$ \frac{\langle \pi^- \Lambda | \Xi^- \rangle} {\langle \pi^0 \Lambda | \Xi^0 \rangle} = \frac { \langle J_\pi=1 , M_\pi=-1 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=-1/2 \rangle} { \langle J_\pi=1 , M_\pi=0 | j_s=1/2 , m_s=-1/2; j_\Xi=1/2 , m_\Xi=1/2 \rangle} =\sqrt{2}. $$
Thus, squaring the amplitude...