Please help with this easy Physics Homework

I basically need to check my answers. I need someone to work out everything and compare it to what I got. It should be easy. Here's the problem:

Find the value of g, acceleration of gravity, in the following situations
A) Earth's mass is triple its actual value, but its radius remains the same
B) Earth's radius is tripled, but its mass remains the same
C) Both the mass and radius of Earth are doubled
D) The acceleration of gravity at a distance of 2r above the Earth's surface

I think you can use the whole F= Gm1m2/r^2 equation. Switch F for mg. The m's cancel out, leaving g = Gm2/r^2. My values for each constant were:
G = Universal Gravitational Constant = 6.67x10^-11 N(m/kg)^2
m2 = mass of Earth = 5.98x10^24 kg
r = radius of Earth = 6.38x10^6 m

My values that I got for each are as follows:
A) -29.40 m/s/s
B) -3.27 m/s/s
C) -9.80 m/s/s (Gravity doesn't change when mass and radius are doubled, apparently)
D)....Now this is a tricky one. I'm not sure how to set this one up. PLEASE HELP.

That's it. PLEASE COMPARE YOUR ANSWERS TO MINE. AND, PLEASE HELP ME ON PART D)!!!! THANK YOU IN ADVANCE!!!

Using M for mass of earth, g = -GM/r² but for neatness I'll omit the minus sign (which simply indicates attraction - as g is directed towards the centre of the earth).

I'll take the 'standard' value of g as 9.8m/s² (surface value); you can work it out from your data if you want.
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A) If M becomes (3M), g' = G(3M)/r² = 3 x GM/r² = 3g = 3x9.8 = 29.4m/s²

No calculation is necessary really. From the equation you can tell tripling the mass triples g.
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B) If r becomes (3r), g' = GM)/(3r)² = (1/9) x GM/r² = g/9 = 9.8/9 = 1.09m/s²

No calculation is necessary really. From the equation you can tell making r 3 times bigger will reduce g by 3² = 9 times. This is the inverse square law.