How much force is required to pull a spring 3.0 cm from its equilibrium position if the spring constant is 3.7 x 103 n/m?

Hooke's law states that for a helical spring the extension is directly proportional to the force applied provided the elastic limit is not exceeded. 
Therefore; F= ke, where k is the spring constant, F is the force and e is the extension.
k = 2700 N/m and e = 3 cm or 0.03 M
therefore, F = 2700 × 0.03
                    = 81 N
Thus, the force required will be 81 N

That depends on what kind of environ meant it is in  in space it has the same force that was applied when the velocity was gained