Answer
$3750\;\rm V/m$ downward, $7500\;\rm V/m$ upward.
Work Step by Step
We know that the electric field is given by
$$E=-\dfrac{dV}{ds}$$
which means that the electric field is the negative of the slope of the $V-s$ graph.
From the given graph, it seems that the potential surfaces are uniformly spaced along the $y$-axis above and below the origin point.
At point 1, which lies between two potential surfaces of 25 V and 100 V.
Whereas the distance between them, on the $y$-axis and above the origin, is 2 cm.
Thus,
$$E_1=-\dfrac{\Delta V}{\Delta y}=\dfrac{100-25}{0.02}$$
$$\vec E_1=-(\color{red}{\bf 3750}\;{\rm V/m})\;\hat j$$
It is pointing downward.
At point 2, which lies between two potential surfaces of 25 V and 100 V.
Whereas the distance between them, on the $y$-axis and below the origin, is 1 cm.
Thus,
$$E_2=-\dfrac{\Delta V}{\Delta y}=\dfrac{100-25}{-0.01}$$
$$\vec E_2= (\color{red}{\bf 7500}\;{\rm V/m})\;\hat j$$
It is pointing upward.
