0.2 M AcOH is ….% dissociation in 0.1 M HCI (K_a of AcOH = 1.8x 10^-5).

$$\begin{gathered} \mathrm{AcOH}\rightleftharpoons {\mathrm{AcO}}^{-}+{\mathrm{H}}^{+}\text{ (weak acid) } \\ \mathrm{C} 0 0 \\ (\mathrm{C}-\mathrm{Cx}) \mathrm{Cx} \mathrm{Cx} \end{gathered}$$
$$\begin{gathered} \mathrm{HCl}\rightleftharpoons {\mathrm{Cl}}^{-}+{\mathrm{H}}^{+}\text{(strong acid) } \\ \text{ Total }\left[{\mathrm{H}}^{+}\right]=\mathrm{Cx}+0.1\approx 0.1\mathrm{M} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{AcO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[\mathrm{AcOH}\right]} \\ 1.8\times {10}^{-5}=\frac{\mathrm{Cx}\times 0.1}{\mathrm{C}(1-\mathrm{x})}\approx 0.1\mathrm{x}\therefore \mathrm{x}=1.8\times {10}^{-4} \\ \mathrm{\%}\text{ dissociation }=1.8\times {10}^{-4}\times 100=0.018\mathrm{\%} \end{gathered}$$