Given that the specific conductance, $k$, is $5 \times 10^{-5} ~S~cm^{-1}$ and the concentration, $C$, is $0.0025~M$, we can find the molar conductivity, $\lambda_m$, as follows:

$$\lambda_m = \frac{k}{C} \times 1000 = \frac{5 \times 10^{-5} \times 10^3}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} = 20 ~S~cm^2~mol^{-1}$$

Next, we find the degree of dissociation, $\alpha$, by dividing $\lambda_m$ by the limiting molar conductivity, $\lambda_m^o$:

$$\alpha = \frac{20}{400} = \frac{1}{20}$$

Finally, we use the formula for the dissociation constant of a weak acid, $K_a$:

$$K_a = \frac{C \alpha^{2}}{1-\alpha} = \frac{0.0025 \times \frac{1}{20} \times \frac{1}{20}}{\frac{19}{20}} = \frac{0.0025}{19 \times 20} = 6.6 \times 10^{-6} = 66 \times 10^{-7}$$

So, the correct answer is 66.