Find the molarity of the filtrate
$$\ce{Cu(NO3)2}+2\ce{NH4OH} =&gt;\ce{Cu(OH)2}(s)+2\ce{NH4NO3} $$
Essentially, we're trying to find the Molarity of $\ce{NH4NO3}$ since the copper filters out(as you stated)

First solve for limiting reagent. $$10mL\ce{Cu(NO3)2}*\frac{1L}{1000mL}*\frac{1.15mol}{1L}=0.0155mol\ce{Cu(NO3)2}$$
$$20mL\ce{NH4OH}*\frac{1L}{1000mL}*\frac{0.85mol}{1L}=0.0170mol\ce{NH4OH}*\frac{1}{2}=0.0085$$


Why do we divide mols of $\ce{NH4OH}$ by 2? Remember that for every molecule of $\ce{Cu(NO3)2}$, there must be 2 molecules of $\ce{NH4OH}$. $$\frac{2\ce{NH4OH}}{\ce{Cu(NO3)2}}$$ for this reason we need to divide the amount of $\ce{NH4OH}$ by 2 to compensate for stoicheometry. 
So actually, $\ce{NH4OH}$ our limiting reagent. 
Since you didn't give my any solubility constants, I'm going to assume 100% of the Copper became  $\ce{Cu(OH)2}(s)$ and all of the ammonium became $\ce{NH4NO3}$. In the real world, perciptates never form at 100%. 


0.0085 mols of $\ce{Cu(OH)2}$ is produced. You want $\ce{NH4NO3}$  Can you use stoichometric ratios to proceed from there? Molarity - (mol/L). Make sure you add up the two volumes to find the Total volume.