For a detailed explanation of the general strategy for solving NMR structure elucidation problems, see problem 662.
Here are the steps I would take to solve this problem:
1. Are there any any hints?
Yes. The sharp IR peak at 1,700 cm^{1} tells you this molecule contains a carbonyl (C=O).
2. How many IHD are there? (also known as DBE or degrees of unsaturation)
C_{4}H_{6}O_{2} is the same as C_{4}H_{6} (oxygens can be ignored) which if fully saturated would be C_{4}H_{10} (from C_{n}H2_{n+2}).
So this compound is missing 4 H's (C_{4}H_{10}  C_{4}H_{6}), which corresponds to 2 IHD.
3. Draw some C_{4}H_{6}O_{2} structures with 2 IHD and eliminate, learn, repeat.
Some things we know so far:

From the IR we know this molecule must have a carbonyl group. This "uses" 1C, 1O, and 1 IHD so we have 3C, 1O and 1 IHD remaining to work with).

What is the the remaining 1 IHD? It's probably not another carbonyl since the ^{13}C NMR shows only one carbonyl peak (~170 ppm). It's probably not an alkene since we don't see any vinyl hydrogen peaks on the ^{1}H NMR (~56 ppm). That leaves a ring.

Another clue that this molecule contains a ring is that we only see CH_{2}'s in the ^{1}H NMR spectrum (all peaks have an integration of 2).

The carbonyl peak on the ^{13}C NMR spectrum is around ~170 ppm, which means it's an oxidation state III carbonyl (ester) rather than an oxidation state II carbonyl (aldehyde or ketone).
Draw a few structures based on these clues and you'll arrive at the correct answer.