This reaction is analogous to the preparation of allyl alcohol from glycerol and oxalic acid at high temperatures. The question as stated with ethylene glycol is unusual, but following the analogous mechanism is the most plausible interpretation.
Step 1: Esterification. Ethylene glycol reacts with oxalic acid to form an intermediate, ethylene monooxalate.
Step 2: Decarboxylation. At high temperatures (around 210°C), the oxalic acid ester decarboxylates (loses CO$_2$) to form a formate ester, ethylene glycol monoformate.
HOOC-COO-CH$_2$CH$_2$OH $\rightarrow$ HCOO-CH$_2$CH$_2$OH + CO$_2$
Step 3: Elimination. Further heating of the formate ester can lead to an elimination reaction. In the analogous reaction with glycerol, this step leads to the formation of a double bond. Here, it is proposed that the ethylene glycol monoformate undergoes elimination to form vinyl alcohol.
HCOO-CH$_2$CH$_2$OH $\rightarrow$ CH$_2$=CHOH + HCOOH (Formic acid)
Vinyl alcohol is an unstable enol and would typically tautomerize to acetaldehyde. However, given the options, vinyl alcohol is the expected primary product from this analogous mechanism.
This pathway is a direct analogy to the well-known synthesis of allyl alcohol from glycerol, which is the likely intent of this problem.