The density of air is 1.3 kg / m3 at sea level. From your knowledge of
air pressure at ground level, estimate the height of the atmosphere.
As a simplifying assumption, take the atmosphere to be of uniform
density up to some height, after which the density rapidly falls to
zero. (In reality, the density of the atmosphere decreases as we go
up.) Height is in km.

Dana
      Why would you want take a square root?

 If you have column of gas of uniform density D
  with height h and area of column A, then


 weight is  W= mass * g  = volume *D *g = A * h * D *g

 and weight/A is the pressure

 So, h would be your 'height in km' IF the case would be realistic.
 
At most, you could stick factor of 2  somewhere 
(assume that pressure goes from max to 0) and 'average is 1/2 of P0'
where P0 is atmospheric pressure at the Earth surface, but your problem
says 'assume uniform' density and that means uniform pressure too.

In reality, profile is is given by  barometric formula, as explained here:

   .. most of the atmosphere is below 100 km (99.9999%) although in
the rarified region above this there are auroras, and other
atmospheric effects.
http://en.wikipedia.org/wiki/Earth's_atmosphere#Thickness_of_the_atmosphere

Normally, I would not suggest 'homework answer', but the 'helpful comment'
and really silly, confusing, problem you were given, is more than any
young person deserves - so ...
 So, to do at least some work,  you should read the whole article given above
and references (and do not argue with the teacher :-)

Hedgie

Homework?

Ian G.

I just need some help/direction.
I was trying to use h =(1.01X10^5)/(1.3kg/m^3)*(9.8)and then take the
square root of h but I believe its not the right answer and I'm stuck
and need some help.

Looks like you're nearly there, although I don't think the square root
is correct (the answer you want is h, taking the square root would
give the wrong units).  I assume you've used Bernoulli's equation,
which simplifies rather nicely for this problem.  It's best that you
figure out velocities and pressures if you haven't already (they're
common-sense).  Hints: velocities - atmosphere is essentially in
static equilibrium; pressures - you've got sea-level pressure, so
what's the pressure at the "end" of the atmosphere (in space)?

Actually the number you should get isn't very accurate, as noted, and
in fact the pressure at that height is still around a third of
sea-level pressure, due to the reduced density at lower pressures.