The acid dissociation constant K_a is related to the degree of dissociation \alpha and the initial concentration C by the approximation K_a \approx C\alpha^2 when \alpha << 1. Given C = 0.1 M and \alpha = 1.34 \times 10^{-2}: K_a = 0.1 \times (1.34 \times 10^{-2})^2 = 0.1 \times 1.7956 \times 10^{-4} = 1.796 \times 10^{-5} \approx 1.80 \times 10^{-5}. Since \alpha = 0.0134 << 1, the approximation (1-\alpha) \approx 1 is valid and introduces negligible error. Option 1.34 \times 10^{-3} is incorrect; it comes from K_a = C\alpha rather than C\alpha^2, forgetting to square the degree of dissociation. Option 1.34 \times 10^{-2} simply equates K_a with \alpha, which is dimensionally and conceptually wrong. Option 2.69 \times 10^{-4} doubles the correct answer, possibly from an error where \alpha is multiplied by 2 instead of squared. This calculation pattern is directly from NCERT Equilibrium and is a standard numerical in JEE Main. Plausibility check: K_a of acetic acid is known to be ~1.8 \times 10^{-5}, confirming our answer is correct in both magnitude and sign.