Q.Which of the following carbocation will be most stable?

Heterolytic cleavage is the unsymmetrical splitting of a covalent bond in which one atom keeps both bonding electrons (becoming an anion) while the other keeps none (becoming a cation); the more electronegative atom is always the one that ends up as the anion, shown with a curved arrow pointing toward it. Unlike homolysis, heterolysis needs no special energetic trigger and readily occurs at room temperature under polar/ionic conditions -- it always requires less energy than homolytic cleavage of the same bond, since it only relocates an intact electron pair rather than genuinely separating one.

Carbocation formation. When the leaving atom/group is more electronegative than carbon (e.g. Br in tert-butyl bromide, $(CH_3)_3C$-Br, hydrolysing to $(CH_3)_3C^+$ + $Br^-$), carbon is left electron-deficient as a carbocation. A carbocation's positively-charged carbon is $sp^2$ hybridised, giving it a planar (trigonal) shape with an empty p-orbital perpendicular to the plane of its three sigma bonds -- this empty orbital is why it carries a positive charge, and its two lobes (above and below the plane) let an attacking nucleophile approach from either face. Carbocation stability rises with alkyl substitution (inductive $+I$ donation plus hyperconjugation from each added alkyl group):

$^+C(CH_3)_3 > {}^+CH(CH_3)_2 > {}^+CH_2CH_3 > {}^+CH_3 \quad (3^\circ > 2^\circ > 1^\circ > \text{methyl})$ …