Answer
$\vec{E}_b > \vec{E}_a > \vec{E}_d > \vec{E}_c$
Work Step by Step
Recall that $\vec{E} = \displaystyle \frac{kq}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$
We can calculate the electric field on each point as:
a. $\displaystyle \vec{E} = \frac{k(q)}{(r)^2} = \frac{kq}{r^2} = \vec{E}$
b. $\displaystyle \vec{E} = \frac{k(2q)}{(r)^2} = 2\frac{kq}{r^2} = 2\vec{E}$
c. $\displaystyle \vec{E} = \frac{k(q)}{(2r)^2} = \frac{1}{4}\frac{kq}{r^2} = \frac{1}{4}\vec{E}$
d. $\displaystyle \vec{E} = \frac{k(2q)}{(2r)^2} = \frac{2}{4}\frac{kq}{r^2} = \frac{1}{2}\vec{E}$
So, from largest to smallest, the electric field at each point ranks
$\vec{E}_b > \vec{E}_a > \vec{E}_d > \vec{E}_c$
