An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1, the magnitude of the car's acceleration is 3.2 m/s2. The magnitude of the car's velocity at the end of stage 2 is 2.0 times greater than it is at the end of stage 1. Find the magnitude of the acceleration in stage 2.

Have you tried setting up the equations for this? Remember \Delta v = a t.

a= 3.2 m/s^2
and
t=2.0
a= 3.2(2.0)
is this what I should be doing?

Not quite, The change in velocity in phase 1 is \Delta v_1 = v_1-v_0 = a_1t . Since v_0 = 0, you have
v_1 = a_1t = 3.2\frac m {s^2} t

The change in velocity in phase 2 is \Delta v_2 = v_2-v_1= a_2t and you know that v_2 = 2v_1.

Can you take it from here?

Could you give me a walk through of all the steps up to the solution I am still unsure how to go about to solve this question.