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Given that, $7$ identical erasers and $4$ kids (distinct).

We have to distribute $7$ erasers in such a way that each kid gets at least one eraser and not more than $3$.

Now, We can give them all one eraser each. So the remaining erasers will be $7-4=3$.

Since no student can get more than $3$ erasers, we can’t give the remaining $3$ erasers to one student only.

Therefore, $2$ cases are possible.

$\textbf{Case 1:}$ Give $1$ eraser each to $3$ students

- $a,b,c,d \quad \text{(Let the name of the kids)}$
- $1,1,1,1 \quad \text{(Already given)}$
- $1,1,1,0$
- $\vdots \;\;\; \vdots \;\; \vdots \;\; \vdots$

Number of possible ways $= \frac{4!}{3!} = 4$ ways

$\textbf{Case 2:}$ Give $1$ eraser to $1$ student and $2$ erasers to other student.

- $a,b,c,d$
- $1,1,1,1 \quad \text{(Already given)}$
- $1,2,0,0$
- $\vdots \;\;\; \vdots \;\; \vdots \;\; \vdots$

Number of possible ways $= \frac{4!}{2!} = \frac{4\times3\times2!}{2!} =12$ ways

$\therefore$ Total number of ways $=4+12=16$ ways.

Correct Answer $:\text{A}$