The upper end of vertical spring of natural length 250 mm is attached to a fixed point. When a small object mass 0.15 kg is attached to the lower end of the spring , the spring stretches to an equilibrium length of 320 mm. calculate the extension of the spring at equilibrium
Calculate the spring constant

What formula do you need for this task ?

What did you calculate? Try Hooke's law. This is a very basic physics experiment that you should know. We will help, not do!

I don't think it is hooke's law. Can you just give me the formula that I need?

The answer should be 70mm but I am not sure which formula to use to get to the answer

Consider a simple helical (https://en.wikipedia.org/wiki/Helix) spring that has one end attached to some fixed object, while the free end is being pulled by a force whose magnitude is https://upload.wikimedia.org/math/8/0/0/800618943025315f869e4e1f09471012.png. Suppose that the spring has reached a state of equilibrium (https://en.wikipedia.org/wiki/Mechanical_equilibrium), where its length is not changing anymore. Let https://upload.wikimedia.org/math/0/2/1/02129bb861061d1a052c592e2dc6b383.png be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched). Hooke's law states that
https://upload.wikimedia.org/math/c/2/c/c2c4a7f57947a786b00bba66d08b109d.pngor, equivalently,
https://upload.wikimedia.org/math/f/2/1/f21d28ad1acbca64d0514c13a9dc963e.png