\begin{align*} \int_{0}^{1}\left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}x\right\rfloor }{3^{k}}\right)^{2}dx & =\int_{0}^{1}\left(\sum_{k=2}^{\infty}\frac{\left\lfloor 2^{k-1}x\right\rfloor }{3^{k-1}}\right)^{2}dx\\ & =9\int_{0}^{1}\left(\sum_{k=2}^{\infty}\frac{\left\lfloor 2^{k-1}x\right\rfloor }{3^{k}}\right)^{2}dx\\ & =18\int_{0}^{\frac{1}{2}}\left(\sum_{k=2}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)^{2}d\frac{x}{2}\\ & =18\int_{0}^{\frac{1}{2}}\left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)^{2}d\frac{x}{2}\tag{(*)}\\ & =18\int_{0}^{1}\left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)^{2}d\frac{x}{2}-18\int_{\frac{1}{2}}^{1}\left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)^{2}d\frac{x}{2}\\ & =\frac{18}{17}\int_{\frac{1}{2}}^{1}\left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)^{2}d\frac{x}{2}\\ & =\frac{18}{17}\int_{0}^{\frac{1}{2}}\left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\left(\frac{x}{2}+\frac{1}{2}\right)\right\rfloor }{3^{k}}\right)^{2}d\frac{x}{2}\\ & =\frac{18}{17}\int_{0}^{\frac{1}{2}}\left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor +2^{k-1}}{3^{k}}\right)^{2}d\frac{x}{2}\\ & =\frac{18}{17}\int_{0}^{\frac{1}{2}}\left\{ \sum_{k=1}^{\infty}\left(\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}+\frac{1}{3}\left(\frac{2}{3}\right)^{k-1}\right)\right\} ^{2}d\frac{x}{2}\\ & =\frac{18}{17}\int_{0}^{\frac{1}{2}}\left\{ \left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)+\frac{1}{3}\cdot\frac{1}{1-\frac{2}{3}}\right\} ^{2}d\frac{x}{2}\\ & =\frac{18}{17}\int_{0}^{\frac{1}{2}}\left\{ \left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)+1\right\} ^{2}d\frac{x}{2}\\ & =\frac{18}{17}\int_{0}^{\frac{1}{2}}\left\{ \left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)^{2}+2\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}+1\right\} d\frac{x}{2}\\ & =\frac{1}{17}\cdot18\int_{0}^{\frac{1}{2}}\left(\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}\right)^{2}d\frac{x}{2}+\frac{18}{17}\int_{0}^{\frac{1}{2}}\left(2\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}+1\right)d\frac{x}{2}\\ & =\frac{17}{17-1}\cdot\frac{18}{17}\int_{0}^{\frac{1}{2}}\left(2\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}+1\right)d\frac{x}{2}\\ & =\frac{9}{8}\int_{0}^{\frac{1}{2}}\left(2\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}+1\right)d\frac{x}{2}\\ & =\frac{9}{4}\int_{0}^{\frac{1}{2}}\sum_{k=1}^{\infty}\frac{\left\lfloor 2^{k}\frac{x}{2}\right\rfloor }{3^{k}}d\frac{x}{2}+\frac{9}{16}\\ & =\frac{9}{4}\sum_{k=1}^{\infty}\frac{1}{3^{k}}\int_{0}^{\frac{1}{2}}\left\lfloor 2^{k}\frac{x}{2}\right\rfloor d\frac{x}{2}+\frac{9}{16}\\ & =\frac{9}{4}\sum_{k=1}^{\infty}\frac{1}{2^{k}3^{k}}\int_{0}^{2^{k-1}}\left\lfloor 2^{k}\frac{x}{2}\right\rfloor d\frac{2^{k}x}{2}+\frac{9}{16}\\ & =\frac{9}{4}\sum_{k=1}^{\infty}\frac{1}{2^{k}3^{k}}\sum_{j=1}^{2^{k-1}-1}j+\frac{9}{16}\\ & =\frac{9}{4}\sum_{k=1}^{\infty}\frac{1}{2^{k}3^{k}}\frac{\left(2^{k-1}-1\right)2^{k-1}}{2}+\frac{9}{16}\\ & =\frac{9}{4}\sum_{k=1}^{\infty}\frac{2^{2k-2}-2^{k-1}}{2^{k+1}3^{k}}+\frac{9}{16}\\ & =\frac{9}{4}\sum_{k=1}^{\infty}\frac{2^{k-3}-2^{-2}}{3^{k}}+\frac{9}{16}\\ & =\frac{9}{4}\sum_{k=1}^{\infty}\left(\frac{1}{2^{2}3}\left(\frac{2}{3}\right)^{k-1}-\frac{1}{2^{2}3}\cdot\frac{1}{3^{k-1}}\right)+\frac{9}{16}\\ & =\frac{9}{4}\left(\frac{1}{2^{2}3}\cdot\frac{1}{1-\frac{2}{3}}-\frac{1}{2^{2}3}\cdot\frac{1}{1-\frac{1}{3}}\right)+\frac{9}{16}\\ & =\frac{9}{4}\left(\frac{1}{2^{2}}-\frac{1}{2^{3}}\right)+\frac{9}{16}\\ & =\frac{9}{32}+\frac{9}{16}\\ & =\frac{27}{32} \end{align*}