Explanation

(b)
For iodoform reaction, we need an oxidising agent which is provided by only $\frac{I_{2}}{KOH},\ i.e.\ ,\ IO^{}^{-}\ ion$.
Hypoiodide ion first oxidises
$CH_{3}CH_{2}OH\ \rightarrow \ CH_{3}CHO$
and then brings about iodination of $CH_{3}CHO\ to\ I_{3}C$ . CHO. Alkaline hydrolysis of $Cl_{3}CHO\$then gives $CHl_{3}.$ The other three reagents do not contain any oxidising species and hence, fail to give iodoform test.