Right choice is (a) 15 Ω
To elaborate: Hybrid parameter h11 is given by, h11 = \(\frac{V_1}{I_1}\), when V2=0.
Therefore short circuiting the terminal Y-Y’, we get,
V1 = I1 ((10||10) + 10)
= I1 \(\left(\left(\frac{10×10}{10+10}\right)+10\right)\)
= 15I1
∴ \(\frac{V_1}{I_1}\) = 15.
Hence h11 = 15 Ω.