1. We need to count integers in $[100,700]$ that are neither divisible by $3$ nor by $4$.

2. First count total natural numbers from $100$ to $700$ inclusive: $700-100+1=601$

3. Let

• $X$ = numbers in $[100,700]$ divisible by $3$
• $Y$ = numbers in $[100,700]$ divisible by $4$

We want: $|A|=601-|X\cup Y|$

4. Count multiples of $3$ in $[100,700]$: $\left\lfloor \frac{700}{3} \right\rfloor-\left\lfloor \frac{99}{3} \right\rfloor=233-33=200$ So, $|X|=200$

5. Count multiples of $4$ in $[100,700]$: $\left\lfloor \frac{700}{4} \right\rfloor-\left\lfloor \frac{99}{4} \right\rfloor=175-24=151$ So, $|Y|=151$

6. Count numbers divisible by both $3$ and $4$, i.e. divisible by $\mathrm{lcm}(3,4)=12$: $\left\lfloor \frac{700}{12} \right\rfloor-\left\lfloor \frac{99}{12} \right\rfloor=58-8=50$ So, $|X\cap Y|=50$

7. By inclusion-exclusion, $|X\cup Y|=|X|+|Y|-|X\cap Y|=200+151-50=301$

8. Therefore, $|A|=601-301=300$

9. Hence the correct option is: $\boxed{\text{A: }300}$

10. Comparison with stored answer: Stored correct answer = A, which matches our result.