The correct option is B ⊕CH=CH2
In the carbocation of the options B, the positive charge is present on sp hybridized carbon atom whereas, in the carbocations of options A,C and D, the positive charge is present on sp3 hybridized carbon atom.
Now sp hybridized carbon atom is more electronegative than sp3 hybridized carbon atom. As the percentage s character increases, the electronegativity increases.
When the positive charge is on the more electronegative carbon atom, the carbocation is less stable.
Thus, the carbocation of option B is less stable.