The way spring design works (roughly) is given: Outside diameter, Wire diameter, Free length, Number of coils, Material... you get a resultant Rate [pounds per inch deflection]. So the more you displace the spring, either tension or compression, the more force it yields. Up to some a maximum force dictated by dimensions (primarily length). What you are asking is the reverse, how much force will your spring inventory give. There is no easy way but just as an example if you needed 5 pounds, you would need to impart that weight on the spring & measure deflection. If that free length & compressed length combination is acceptable to your design, you have your winner. Just realize spring A may yield 5 pounds in 0.1" deflection vs. spring B may yield 5 pounds in 0.6" deflection. They both do the same job, but may have radically different dimensions.
If your springs are out of range then its best to go to some kind of product table or online calculator to get you in the ball park. Attaching table showing variation just around an arbitrary picked OD. When you double & triple OD & wire thickness (like your collection) the forces will vary significantly.
https://www.thespringstore.com/spring-calculator.html
But remember this gives you force (like pounds). For a relief valve you need to it to open at some max pressure. So that involves (I assume) also knowing area, like a ball valve seat OD. Pressure = force / area.
I hope you have a relief valve design that specifies everything including spring, ball, seat dimensions & thread. These are all critically important. Don't underestimate a 'toy model' boiler because any compressed gas pack a lot of energy that you don't want to go boom. Especially hot boom. There must be a bunch of designs in the model engineering (live steam) community. This is roughly what they look like (note the pressure setting & number of valves).
http://www.americanmodeleng.com/id6.html