For the below given cyclic hemiacetal X, the correct pyranose structure is

To convert a cyclic hemiacetal from a Fischer projection to a Haworth pyranose structure, standard conversion rules are followed.

A six-membered pyranose ring is drawn with the ring oxygen at the top right.

Conversion rule:

$$\mathrm{Left\ side\ in\ Fischer \rightarrow Up\ in\ Haworth}$$

$$\mathrm{Right\ side\ in\ Fischer \rightarrow Down\ in\ Haworth}$$

Applying this rule:

At $$\mathrm{C1}$$ (anomeric carbon):

The $$\mathrm{-OH}$$ group is on the left side.

Therefore:

$$\mathrm{-OH \ points\ Up}$$

At $$\mathrm{C2}$$:

The $$\mathrm{-OH}$$ group is on the left side.

Therefore:

$$\mathrm{-OH \ points\ Up}$$

At $$\mathrm{C3}$$:

The $$\mathrm{-OH}$$ group is on the right side.

Therefore:

$$\mathrm{-OH \ points\ Down}$$

At $$\mathrm{C4}$$:

The $$\mathrm{-OH}$$ group is on the left side.

Therefore:

$$\mathrm{-OH \ points\ Up}$$

At $$\mathrm{C5}$$:

The ring oxygen is on the right side, indicating a $$\mathrm{D}$$-sugar.

For $$\mathrm{D}$$-sugars in Haworth form:

$$\mathrm{-CH_2OH \ points\ Up}$$

Therefore, the correct Haworth pyranose structure has:

$$\mathrm{C1:\ Up}$$

$$\mathrm{C2:\ Up}$$

$$\mathrm{C3:\ Down}$$

$$\mathrm{C4:\ Up}$$

and

$$\mathrm{-CH_2OH \ group\ Up}$$