Official solution
I comes out around 1.536 which is not satisfied by any given options.
\[\int\limits_{\pi /4}^{\pi /3} {{{8x - 2x} \over x}dx > I > \int\limits_{\pi /4}^{\pi /3} {{{8\sin x - 2x} \over x}dx} } \]
\[{\pi \over 2} > I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x} \over x} - 2} \right)dx} \]
\[{{\sin x} \over x}\] is decreasing in \[\left( {{\pi \over 4},{\pi \over 3}} \right)\] so it attains maximum at \[x = {\pi \over 4}\]
\[I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x/3} \over {x/3}} - 2} \right)dx} \]
\[I > \sqrt 3 - {\pi \over 6}\]