We have,

$3{\cot }^{2}x+8\cot x+3=0$

$\therefore \cot x=\frac{-4\pm \sqrt{7}}{3}$

Both roots are negative.

$\therefore \frac{\pi }{2}<x_1,x_2<\pi$

$\therefore \pi <x_1+x_2<2\pi$.

Product of roots $=\cot x_1\cot x_2$

 $=1$

$\Rightarrow \cot x_1=\frac{1}{\cot x_2}$

$\Rightarrow \cot x_1=\tan x_2$

$\Rightarrow \tan \left(\frac{3\pi }{2}-x_1\right)=\tan x_2$

$\Rightarrow \frac{3\pi }{2}-x_1=x_2$

$\Rightarrow x_1+x_2=\frac{3\pi }{2}$

$\Rightarrow k=\frac{3\pi }{2}$

$\Rightarrow \frac{k}{\pi }=\frac{3}{2}=1.50$