An alkene (X) on ozonolysis followed by reduction gives following...

Ozonolysis (followed by reductive workup with $Zn/H_2O$) cleaves $C=C$ bonds and converts each alkene carbon into a carbonyl. We can reverse-engineer the parent alkene from the fragments obtained.

Analysing the products:
$HCHO$ $(2\text{ moles})$: Two equivalents of formaldehyde indicate the presence of two terminal or exocyclic $=CH_2$ groups in the parent molecule.

$CH_3-CO-CO-CH_3$ (biacetyl): To get this discrete fragment, the parent alkene must contain the structural unit $-C(CH_3)=C(CH_3)-$, i.e., an endocyclic double bond in which both sp$^2$ carbons bear methyl substituents.

$H-CO-CO-CHO$ (mesoxaldehyde): This arises from a fragment like $=CH-C(=CH_2)-CH=$, where cleavage of the adjacent $C=C$ bonds generates three consecutive carbonyl groups (and also contributes to the formaldehyde count from the exocyclic $=CH_2$).

Evaluating the options:
The most diagnostic product is biacetyl, so the parent structure must contain the $-C(CH_3)=C(CH_3)-$ unit.

Option $(1)$: One methyl group is on the endocyclic double bond and the other methyl is on a separate sp$^3$ carbon. Cleavage cannot give biacetyl.

Option $(2)$: Only a single methyl group is present, so biacetyl cannot be formed.

Option $(4)$: The methyl groups lie on opposite ends of the ring (across a $1,4$-diene-type arrangement). Ozonolysis would place these methyls in different fragments, so biacetyl cannot be formed.

Option $(3)$: The left side of the ring contains an endocyclic double bond in which both carbons carry methyl groups, giving exactly the required $-C(CH_3)=C(CH_3)-$ arrangement. Cleavage of this bond together with the adjacent ring double bonds releases a discrete $CH_3-CO-CO-CH_3$ fragment. The right side of the molecule carries exocyclic $=CH_2$ groups, which on cleavage produce the two moles of $HCHO$, while the intervening carbons give rise to $H-CO-CO-CHO$.

Hence, the correct structure that yields all three ozonolysis products is option $(3)$.