Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
It is obvious that the electric potential $V$ is always positive everywhere which means that the two charges must be positive.
As a proof, we can see at $a\lt x\lt b$, the electric potential is also positive.
$$\color{blue}{\bf [b]}$$
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,
$$\boxed{\dfrac{|q_a|}{|q_b|}=1} $$
$$\color{blue}{\bf [c]}$$
Since the two charges are positive, the elctric field at $x\gt b$ is pointing to the right, at $x\lt a$ is pointing to the left, and at $a\lt x\lt b$ is a vector sum of the two electric fields which must zero at the middle distance between both charges.
See the figure below.
