Principle for $S_N1$ reaction

An $S_N1$ reaction proceeds through formation of a carbocation in the rate-determining step. So, a halide will show $S_N1$ only if loss of halide ion can generate a reasonably stable carbocation.

Examine option (A)

$\mathrm{H_2C=CH-CH_2Cl}$

This is allyl chloride. On ionization, it forms an allylic carbocation:

$\mathrm{H_2C=CH-CH_2^+} \leftrightarrow \mathrm{^+CH_2-CH=CH_2}$

This carbocation is resonance stabilized, so (A) can undergo $S_N1$.

Examine option (B)

$\mathrm{CH_3-CH=CH-Cl}$

Here chlorine is attached directly to a double-bonded carbon; this is a vinylic halide.

If $Cl^-$ leaves, the carbocation formed would be a vinyl carbocation, which is highly unstable:

$\mathrm{CH_3-CH=CH^+}$

Therefore, vinylic halides do not undergo $S_N1$ reaction.

Conclusion from given answer choices

Since (A) shows $S_N1$ and (B) does not, the only option consistent with this is:

$\boxed{\text{(B) only}}$

which corresponds to Option C.

Comparison with stored correct answer

Stored correct answer = C

Derived answer = C

Hence, they agree.