The correct representation in six membered pyranose form for the following sugar [X] is:
Sugar [X] has the Fischer projection: CHO, HO-H, HO-H, H-OH, H-OH, $$H_2COH$$

Step 1 - Identify the open-chain sugar.

The Fischer projection is
CHO (C1)
HO-H (C2, OH on left)
HO-H (C3, OH on left)
H-OH (C4, OH on right)
H-OH (C5, OH on right)
CH2OH (C6)

For an aldohexose, read the configuration from C2 to C5 (R = OH right, L = OH left).
Sequence: L L R R.

This pattern matches D-mannose (D series because the lowest chiral centre, C5, has the -OH on the right).

Step 2 - Form the pyranose ring.

The -OH on C5 attacks the aldehydic C1 to give a six-membered hemiacetal (a pyranose).
Hence the ring will contain O (from C5 oxygen) and the atoms C1→C5.

Step 3 - Place substituents in the Haworth projection.

Convention: for a D-sugar drawn with the anomeric carbon (C1) at the right-hand side of the ring, groups that were on the RIGHT in the Fischer appear BELOW the ring, while groups that were on the LEFT appear ABOVE the ring.

Apply this to D-mannopyranose:

C1 (anomeric) α-OH ↓ or β-OH ↑ (both possible)
C2 OH was left → OH ↑
C3 OH was left → OH ↑
C4 OH was right → OH ↓
C5 CH2OH is fixed above the plane for D-sugars → CH2OH ↑

Step 4 - Match with the given options.

The only option that shows
• both C2 and C3 hydroxyl groups ABOVE the ring,
• the C4 hydroxyl group BELOW the ring,
• the C5 CH2OH ABOVE the ring, and
• a correct α/β anomer at C1,
is Option B.

Therefore the correct six-membered (pyranose) representation of sugar [X] is Option B, which depicts D-mannopyranose with the required stereochemistry.

Answer: Option B which is: D-mannopyranose.