what is the cross aldol product of ethanal and acetone(major product)?
As pointed out by Waylander, the question essentially wants students to go for the most stable product, or thermodynamic product. This is sort of like a well-understood thing for aldol product determination questions. Why? Well... Simply because there is usually only 1 thermodynamic product while there are several possible kinetic products. It is therefore hard to argue that the kinetic product drawn is really the major product under kinetic control. Under kinetic control, you may just get a mixture of products with nothing formed in significantly higher quantity. 

Based on this understanding, we can go on to determine the thermodynamic product. This reaction is a very straightforward aldol reaction due to there being only two possible reaction pathways since one reactant is a symmetrical ketone while the other is an aldehyde. So there really are only two possible products...

One possible product comes from aldehyde as a the electrophile and acetone behaving as the nucleophile. Upon dehydration, we obtain A, which is a ketone and has a di-substituted conjugated $\ce {C=C}$ double bond. The other possible product comes from aldehyde as the nucleophile and acetone as the electrophile. Upon dehydration, we obtain B, which is still an aldehyde but it has a tri-subustituted $\ce {C=C}$ bond.



Which one is the thermodynamic product? Usually, we will just look at the substitution of the conjugated double bond formed. Very clearly, the kinetic product must be A since the more nucleophilic enolate from acetone attacks the more electrophilic aldehyde. However, is B really more stable? No, it isn't. The explanation presented below is based on qualitative molecular orbital theory. 

Well... The key differences between A and B are:


A, being a ketone, has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=O \pi^*}$ MO, compared to the B
B instead has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=C \pi^*}$ MO, compared to A


The former hyperconjugative interactions are stronger since the $\ce {C=O \pi^*}$ MO is lower in energy and thus more energetically-accessible to the $\ce {C-H \sigma}$ MO. This is due to the higher electronegativity of the oxygen atom. Thus, we would expect the thermodynamic stability of A to be higher than that of B.

To provide even more concrete substantiation for the higher stability of A over B, Karsten Theis has kindly provided a source of data which indicates the Gibbs free energy of formation of the two compounds. As stated on chemeo, here and here, A has $\Delta G_f = \ce {-57.48 kJ/mol}$ while B has $\Delta G_f = \ce {-36.63 kJ/mol}$. This very clearly indicates that A has higher stability than B.

Thus, A is the answer.