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 $$x = {\pi \over 4}$$

$$I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x/3} \over {x/3}} - 2} \right)dx} $$

$$I > \sqrt 3 - {\pi \over 6}$$