Calcium carbide (cac2) reacts with water to form acetylene (c2h2) and ca(oh)2.
from the following enthalpy of reaction data and data in appendix c, calculate δh∘f for cac2(s):
cac2(s)+2h2o(l)→ca(oh)2(s)+c2h2(g)δ h∘=−127.2kj
δh∘f = kj?

The answer is -60.57. With appropriate sig figs it becomes -60.6 KJ. 

I think you were on the right track. However you need to multiply H2O by its coefficient from the balanced equation. Then you also need to keep in mind the signs. 

Maybe it's easier if I write what I did out: 

2H2O > 2H2 +O2-H= 2* -285.83 
Ca + O2 + H2 > Ca(OH)2H= -986.2 
2C + H2 > C2H2H= 226.77 

The above H's stand for standard enthalpy of formations. These can be found in textbook appendix. Notice the negative infront of the enthalpy (H) for H2O. This is to remind me/you that the heat lost is gained in the rxn. 

So then you add them up. 226.77 - 986.2 + (2*285.83) = -187.77 

Add back the total enthalpy that is given in the question -187.77+127.2 = -60.57 

If signs cross you up find a way to remember that works for you.