Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.
A more combinatorial argument: The number of ways to write $n$ as a sum of $k$ nonnegative integers is
and a discussion can be found here.
So, in this case, the number of ways that $a_1+a_2+a_3=3$ (without restrictions) is
This, however, counts too many possible sums. Suppose that we take too many from box $1$, this means that we take at least $2$ from box $1$. In this case, we can write $a_1=2+b_1$ where $b_1$ is nonnegative. Then, the initial sum becomes $b_1+a_2+a_3=1$. Using the same formula, this results in
impossible ways. Continuing, there is no way to take too many objects from the second box, but it is possible to take too many objects from box $3$. In this case, one must take $3$ objects from box $3$, so we write $a_3=3+b_3$ where $b_3$ is nonnegative. This results in the equation $a_1+a_2+(3+b_3)=3$. There are
ways for this sum to occur. We should now use the inclusion/exclusion principle to see if we've over-counted. This could happen if we take more than $1$ item from box $1$ and more than $2$ items from box $3$. Then, we have $(2+b_1)+b_2+(3+b_3)=3$, but this has no solutions as a sum of nonnegative integers cannot be negative.
Therefore, out of the original $10$ possibilities, $3+1=4$ are impossible, leaving the $6$ that we've found.
A dynamic programming-type solution: Let $N(b,s)$ be the number of ways to use the first $b$ boxes to sum to $s$. Also, write $m_i$ for the number of objects in box $i$. In your case:
Then, the values in the second box can be computed as follows:
Using this formula:
Continuing for the third column,
We are interested in the value $N(3,3)=6$.