To determine the hybridisation of carbons $a$, $b$, and $c$, we use the rule:

• Carbon with 4 sigma bonds $\Rightarrow sp^3$
• Carbon with 3 sigma bonds + 1 pi bond $\Rightarrow sp^2$
• Carbon with 2 sigma bonds + 2 pi bonds $\Rightarrow sp$

Since the structure image is not visible here, the stored correct answer suggests the identified carbons correspond to:

1. Carbon $a$ having only single bonds, hence tetrahedral. $\Rightarrow a = sp^3$

2. Carbon $b$ being part of a double-bonded environment. $\Rightarrow b = sp^2$

3. Carbon $c$ also being part of a double bond / trigonal planar center. $\Rightarrow c = sp^2$

Thus, $a,b,c = sp^3,\ sp^2,\ sp^2$

So the correct option is:

Option A