0.1 M solution of CH₃COOH should be diluted to ...times so that pH is doubled

$$\begin{gathered} \mathrm{pH}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{a}}-\log 1\right]=\frac{{\mathrm{pK}}_{\mathrm{a}}}{2} \\ \left.{\mathrm{pH}}^{\mathrm{\prime }}\text{ (twice of }\mathrm{pH}\right)=\underset{1}{\mathrm{p}}{\mathrm{K}}_{\mathrm{a}} \\ \therefore {\mathrm{pK}}_{\mathrm{a}}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{a}}-\mathrm{logC}\right] \\ -\mathrm{logC}={\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}} \\ \mathrm{C}={\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5}\mathrm{M} \\ \therefore \text{ Dilution }=\frac{1}{\mathrm{C}}=5.55\times {10}^4\text{ times } \end{gathered}$$