$\mathrm{HX}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}$

$$\begin{aligned} &0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad \text { Not justified }\\ &\begin{aligned} & \Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3} \\ & \mathrm{~K}_{\mathrm{a}}=0.01 \alpha^2=10^{-8} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { On dilution let final concentration of } \mathrm{HX}=\mathrm{c} \mathrm{M}\\ &\mathrm{Hx}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \end{aligned}$$

$\mathrm{C}(1-\alpha) \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha$

$$\begin{aligned} \Rightarrow & C \alpha=10^{-6} \quad\text{..... (1)}\\ & \frac{C \alpha^2}{1-\alpha}=\mathrm{K}_{\mathrm{a}}=10^{-8} \quad\text{..... (2)}\\ \Rightarrow & \frac{10^{-6} \alpha}{1-\alpha}=10^{-8} \end{aligned}$$

Data given is inconsistent & contradictory. This should be bonus.