$\begin{aligned} &\begin{aligned} & 10 \sin ^4 \theta+15 \cos ^4 \theta=6 \\ \Rightarrow & 10 \sin ^4 \theta+10 \cos ^4 \theta+5 \cos ^4 \theta=6 \\ \Rightarrow & 10\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta\right]+5 \cos ^4 \theta=6 \\ \Rightarrow & 10-20\left(1-\cos ^2 \theta\right) \cos ^2 \theta+5 \cos ^4 \theta=6 \end{aligned}\\ &\text { Let } \cos ^2 \theta=x\\ &10-20\left(x-x^2\right)+5 x^2=6 \end{aligned}$ $\begin{aligned} \Rightarrow & 25 x^2-20 x+4=0 \\ & (5 x-2)^2=0 \Rightarrow x=\frac{2}{5} \\ \Rightarrow & \cos ^2 \theta=\frac{2}{5} \Rightarrow \sin ^2 \theta=\frac{3}{5}, \\ & \sec ^2 \theta=\frac{5}{2}, \operatorname{cosec}^2 \theta=\frac{5}{3} \end{aligned}$ $\begin{aligned} \frac{27 \operatorname{cosec}^6 \theta+8 \sec ^6 \theta}{16 \sec ^8 \theta} & =\frac{27\left(\frac{5}{3}\right)^3+8\left(\frac{5}{2}\right)^3}{16\left(\frac{5}{2}\right)^4} \\ & =\frac{5^3+5^3}{5^4}=\frac{2.5^3}{5^4}=\frac{2}{5} \end{aligned}$