Roughly, although you don't have to try every possibility. It's clear from the table that you have to either have $g(b)=1$ and $g(c)=2$ or $g(b)=2$ and $g(c)=3$, as $b,c$ and $2,3$ are the elements that give you the same answer no matter what you multiply by, and then checking how they multiply with $a$ and $1$ tells you which way round they have to be.

Ok sir. I get it. Thank you so much for your help.

That's good, because that comment has very confusing typos in it! And I can't edit it now. I meant "...$g(b)=2$ and $g(c)=1$, as $b,c$ and $1,2$...", and "...checking how they multiply with $a$ and $3$..."

Sir, I have another doubt. These two tables are of same size. But what if both are of different size. For eg. $+_4$ ([0],[1],[2],[3]) and B={0,1} with operation + ?

An isomorphism is in particular a bijection, so two algebraic structures with different cardinalities cannot be isomorphic.

@MattPressland But, it has been given that, by taking g([0])=g([2])=0 and g([1])=g([3]) = 1 they have proved it as isomorphic sir.

That's not an isomorphism (it has no inverse). It is, however, a homomorphism. You can define homomorphisms easily by defining them on a generator and extending. In this case, [1] generates the integers mod 4, by which I mean everything in this group is a sum of [1]s. Then if you define g([1])=x, you have to define g([2])=g([1]+[1])=g([1])+g([1])=x+x, and so on.

I should say "defining them on a generating set", there will sometimes be more than one generator. You have to check also that any relations (equations involving the generators) are satisfied by the images of these generators under g. (If this isn't clear, don't worry, it probably isn't important).

oh...sorry sir. I read the question wrongly. It's given as homomorphic only and not isomorphic.

No problem. (As an aside, "homomorphic" isn't really used as a term, because you generally expect algebraic structures to always have at least one homomorphism between them, such as mapping everything to the identity).

Can you kindly explain how to prove it by taking values as 0 and 1? should I replace them by 0 and 1?

I'm not sure what you mean - if you want to check that the map they've given you is a homomorphism, you just need to check that g(x+y)=g(x)+g(y) for each pair of x and y in {[0],[1],[2],[3]}.

I have to check whether binary operation {0,1} is a homomorphic image of ($Z_4,+_4)$, Sir.

congruence modulo 4

Right. So if you check that the map g you defined above is a homomorphism (and that it's surjective, but this is clear) then you'll have done that. Asking where such a homomorphism comes from is a good question, but one that's probably easier to answer yourself through practice.

ok sir. I get it. Thank you so much for your help, Sir.

No problem.