1. Use the beat frequency relation

When two forks of frequencies $f_1$ and $f_2$ are sounded together, the number of beats per second is

$|f_1-f_2|$

Initially, $4$ beats per second are heard, and $f_1=200\text{ Hz}$. So,

$|200-f_2|=4$

Hence the original frequency of fork $2$ can be either

$f_2=204\text{ Hz} \quad \text{or} \quad f_2=196\text{ Hz}$

2. Effect of attaching tape to fork 2

Attaching tape to a tuning fork increases its mass, so its frequency decreases.

After attaching tape, the beat frequency becomes $6$ beats per second. This means the new frequency of fork $2$ moves farther away from $200\text{ Hz}$ than before.

3. Check both possibilities

• If originally $f_2=204\text{ Hz}$, then after attaching tape its frequency decreases, say to something less than $204$. The difference from $200$ would then decrease from $4$ or at most could become smaller before crossing $200$. It cannot directly explain an increase to $6$ beats unless we know it crosses below $200$, which is not the standard intended interpretation.

• If originally $f_2=196\text{ Hz}$, then after attaching tape its frequency decreases further, below $196$. Then its difference from $200$ increases from $4$ to $6$, which is perfectly consistent.

So the original frequency of fork $2$ was

$\boxed{196\text{ Hz}}$

4. Option check

• A: $202\text{ Hz}$ ❌
• B: $200\text{ Hz}$ ❌
• C: $204\text{ Hz}$ ❌
• D: $196\text{ Hz}$ ✅

Therefore, the correct answer is D.