1. Let the three children receive $x_1, x_2, x_3$ apples respectively.

   Since the apples are identical and each child must get at least $2$ apples,

   \quad x_1,x_2,x_3 \ge 2.$$

2. Remove the minimum required apples.

   Let $y_1=x_1-2,\quad y_2=x_2-2,\quad y_3=x_3-2,$ so that $y_1,y_2,y_3 \ge 0.$

   Then $y_1+y_2+y_3=21-6=15.$

3. Now we need the number of non-negative integer solutions of $y_1+y_2+y_3=15.$

   Using the stars and bars formula, the number of solutions is $\binom{15+3-1}{3-1}=\binom{17}{2}.$

4. Compute: $\binom{17}{2}=\frac{17\cdot 16}{2}=136.$

5. Therefore, the required number of ways is $\boxed{136}.$

6. Checking options:

   • A: $130$ ❌
   • B: $136$ ✅
   • C: $142$ ❌
   • D: $406$ ❌

Hence, the correct option is B.