Question

An electron is released from rest in a uniform electric field. the electron accelerates vertically upward, traveling 4.50m in the first 3.00μs after it is released.

Answer 1

If you are looking for the strenght of the electric field, then what you need to do is the following calculation:
E=F/q, and F=ma,
thus E=(ma)/q=(m· 2· x)/(t 2· q) = [(9.11×10-31kg)·2·(4.5m)] / [(3×10-6s)2(1.6×10-19C )]= 5.69 N/C 
That is the strenght. Now that you know that the force is up and that the electron is negative then the projectile motion principles must be used
.