H-NMR with 10H integration?
I will ignore the IR as I always do and work only with the NMR.


The ten aryl protons are probably two phenyl groups ($\ce{C6H5-{}}$). That leaves $\ce{C4H6O2}$ and takes care of eight of the nine double-bond equivalents.
The two triplets are very likely (examining the coupling constant would confirm that) part of a $\ce{-CH2-CH2-{}}$ fragment. On either side of these is either an oxygen or a quarternary carbon as there are no additional couplings.
The final singlet points to a $\ce{-CH2-{}}$ group with the same constraints as above.
We still have $\ce{CO2}$ and one double-bond equivalent to take care of. One functional group particularly suggests itself. Then, we have to play connect the fragments.

   $\ce{CO2}$ is likely an ester functionality since we do not have a carboxylic hydrogen to spare.


   One of the two triplet $\ce{CH2}$ is shifted downfield rather strongly. This suggests that it is directly bound to the ester group oxygen, giving $\ce{-CH2-CH2-OC(=O)-{}}$.


   Both of the remaining $\ce{CH2}$ groups will be benzylic but the singlet is more deshielded than the triplet. That suggests that the singlet $\ce{CH2}$ is not only benzylic but also &#x3b1; to the ester.


   Therefore, the final result is $\ce{C6H5-CH2-CH2-OC(=O)-CH2-C6H5}$.