Step 1: Focus on why the first substitution goes where it does.
The $-OH$ group on phenol pushes electron density strongly into the ring by resonance, most of all at the ortho and para positions. Working at a low temperature and in a comparatively non polar solvent limits the reaction to a single, clean substitution rather than multiple bromination, and the para position is favored since it avoids crowding next to the bulky $-OH$ group.
Step 2: Track what the sodium and ether step is doing.
Treating the brominated intermediate with sodium metal in dry ether is a reductive step: it strips out the halogen that was just introduced and replaces it with a new bond involving oxygen, converting that position into a second hydroxyl bearing carbon.
Step 3: Put the two steps together.
Step one plants a substituent para to the original $-OH$; step two then converts that same para position into a hydroxyl group as well, so the ring ends up with two $-OH$ groups directly across from each other.
Step 4: Name the resulting structure.
A benzene ring carrying two $-OH$ groups at the 1 and 4 positions is hydroquinone, also called 1,4-dihydroxybenzene.
Step 5: Rule out the alternatives.
An ether linkage, a fused ring system, or a biphenyl type product would all need a completely different set of reagents than the ones given, so none of them fit this sequence.
Step 6: Final answer.
\[ \boxed{\text{Hydroquinone (1,4-dihydroxybenzene)}} \]
