a) The acceleration of the proton is 
b) The time required to reach the given velocity is 
Explanation:
a)
This is a motion at constant acceleration, so we can use the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the proton in this problem, we have:
is the final velocity
is the initial velocity (it starts from rest)
is the distance covered
Solving for a, we find the acceleration:

b)
For this part, we can use the following suvat equation instead:

where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken for the velocity to change from u to v
We have here the following data:
is the final velocity
is the initial velocity (it starts from rest)

Solving for t, we find

Learn more about accelerated motion here:
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