Answer
$ \frac{Q_1}{Q_2}=\frac{1}{4}$
Work Step by Step
The charges are either both positive or both negative so that the electric fields are oppositely directed and add to zero. Their magnitudes must be equal.
$$E_1=E_2$$
$$k\frac{Q_1}{(\mathcal{l}/3)^2}= k\frac{Q_2}{(2\mathcal{l}/3)^2}$$
$$ 9Q_1=\frac{9Q_2}{4}$$
$$ \frac{Q_1}{Q_2}=\frac{1}{4}$$