One dozen of big marbles and small marbles is 132 gram. If one big marbles is 3 gram heavier than one small marbles, then specify the possibilities of how many are the big marbles and the small marbles.
I got $x+y=12$, $ax+by=132$, and $a=3+b$ with $x$ is how many big marbles, $y$ is small marbles, $a$ and $b$ are their mass. I don't know what to do next.
Let's express everything in terms of $b$. We have $a=b+3$. Putting that in the second equation gives $(b+3)x+by=132$, which is $3x+b(x+y)=132$, but from the first equation $x+y=12$, so $3x+12b=132$, so $x=44-4b$. Then $y=12-x=4b-32$.
Now, we need $x\ge0$, so $b\le11$, and $y\ge0$, so $b\ge8$. So, $8\le b\le11$. Further, $x$ and $y$ need to be whole numbers, so $4b$ must be a whole number, so $b$ must be one of the 13 numbers $8,33/4,17/2,\dots,11$. For each such $b$, you can use the formulas already derived to work out $x$, $y$, and $a$.
EDIT: The upshot is that $x$ can be any whole number between zero and twelve, inclusive. Then $y=12-x$, $b=11-(x/4)$, and $a=14-(x/4)$.
MORE EDIT: If you read the question to imply there must be at least one marble of each of the two types, then $x$ can be any whole number between one and eleven, inclusive (and $y,b,a$ are given by the formula above).