Well, if you find the approximate change in V, using linear approximation, you'll get
f(x) = 6x^3 = V
f(x+h) = f(x) + hf'(x) x = 3, h = 2/100, f'(x) = 18x^2
f(3.02) = 6(3^3) + (2/100)(18*3^2)
f(3.02) = 6(27) + (2/100)(162)
f(3.02) = 162 + 3.24
f(3.02) = 165.24
So, the change in V from 3 to 3.02 is 165.24 - 162 = 3.24
If you wanted to know the actual change, you have to draw a cubic graph. Now, you realise that an x^3 graph is always increasing, and that it's derivative is always getting greater with a greater x.
So, f'(3.02) > f'(3)
Since we approximated using the derivative at 3, not 3.02 (which is the point of linear approximation anyway), the change in V will be lower than the actual change.
You can always check it by doing:
V(3.02) - V(3) = 165.2616 - 162 = 3.2616