Let:

• $P(\text{I hits}) = 0.3$
• $P(\text{II hits}) = 0.2$

The second plane bombs only if the first plane misses.

So first compute the probability that plane I misses: $P(\text{I misses}) = 1 - 0.3 = 0.7$

For the target to be hit by the second plane, two things must happen:

• plane I misses, and
• plane II hits.

Therefore, $P(\text{target hit by II}) = P(\text{I misses}) \cdot P(\text{II hits})$

Substitute the values: $P(\text{target hit by II}) = 0.7 \times 0.2 = 0.14$

So the required probability is: $\boxed{0.14}$

Checking options:

• A: $0.2$ ❌
• B: $0.7$ ❌
• C: $0.06$ ❌
• D: $0.32$ ❌

Hence, none of the given options matches the correct value.

Comparison with stored correct answer:

• Stored correct answer: D ($0.32$)
• Derived answer: $0.14$

Therefore, the stored answer appears to be incorrect.