Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric field is a vector quantity, and we can see from the given figure that the electric field at $x\gt b$ is negative and is pointing left. This means that the charge at point $b$ is a $\text{negative charge}$.
And at $x \lt a$, we can see from the given figure that the electric field is also pointing left. Hence, the charge at point $a$ is a $\text{positive charge}$.
As proof, we can see at $a\lt x\lt b$, that the electric field is pointing upward which is the sum of the two electric fields from the two charges. The electric field from $b$ points right toward it and the electric field from $a$ points right away from it.
$$\color{blue}{\bf [b]}$$
It seems that the two charges are equal in magnitude due to the symmetry of the given figure at the middle point between the two charges.
Hence,
$$\boxed{\dfrac{|q_a|}{|q_b|}=1} $$
$$\color{blue}{\bf [c]}$$
See the figure below.
