a) 
The gravitational field strength near the surface of the asteroid is given by:

where
G is the gravitational constant
M is the mass of the asteroid
R the radius of the asteroid
h is the distance from the surface
Substituting the data of the asteroid:
is the mass
is the radius of the asteroid
is the distance from the surface
We find

b) i) 
The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):

So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:

where
s = 3 km = 3000 m is his displacement to reach the surface
u = 0 is his initial velocity
t is the time
Solving for t,

b) ii) 
Again, we can use another suvat equation:

where
v is the final velocity
u is the initial velocity
g is the acceleration of gravity
t is the time
Since we have
u = 0


The velocity of the astronaut at the surface will be

b) iii) 175 years
The duration of one year here is

And the time it takes for the astronaut to reach the surface of the asteroid is

Therefore, to find the number of years, we just need to divide the total time by the duration of one year:

So, the astronaut will take 175 years to reach the surface.