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The iodoform test is given by compounds containing $\ce{-(C=O)CH3}$ group or compounds which can be oxidized to such. $$\ce{ CH3CH(OH)CH2CH3}$$
Is a 2 degree alchohol and even though it were to be oxidized it would not contain a $\ce{-(C=O)CH3}$ group but according to a question asked by JEE in 1997 it as a matter of fact does give the test.
HOW?
The substrate you have provided (2-butanol) would indeed give a positive test.
The iodoform test is performed in presence of iodine and potassium or sodium hydroxide, which first oxidizes the aforesaid compound to 2-Butanone.
Mechanism (taken from this Wikipedia page) which shows $\ce{BrO-}$ instead of $\ce{IO-}$, but would be executed in a similar fashion:
Thus 2-Butanone responds to iodoform test successfully.
References:
