1. Let $s=\sin^2\theta,\qquad c=\cos^2\theta$ Then $s+c=1$ and the given equation becomes $10s^2+15c^2=6$

2. Substitute $s=1-c$: $10(1-c)^2+15c^2=6$ $10(1-2c+c^2)+15c^2=6$ $10-20c+10c^2+15c^2=6$ $25c^2-20c+4=0$

3. Solve the quadratic: $25c^2-20c+4=(5c-2)^2=0$ So, $c=\cos^2\theta=\frac{2}{5}$ Hence, $s=1-c=1-\frac{2}{5}=\frac{3}{5}$

4. Now simplify the required expression: $\frac{27\csc^6\theta+8\sec^6\theta}{16\sec^8\theta}$ Using $\csc^6\theta=\frac{1}{\sin^6\theta}=\frac{1}{s^3},\qquad \sec^6\theta=\frac{1}{\cos^6\theta}=\frac{1}{c^3},\qquad \sec^8\theta=\frac{1}{c^4}$ we get $\frac{27/s^3+8/c^3}{16/c^4}$ $=\frac{c^4}{16}\left(\frac{27}{s^3}+\frac{8}{c^3}\right)$ $=\frac{27c^4}{16s^3}+\frac{8c}{16}$ $=\frac{27c^4}{16s^3}+\frac{c}{2}$

5. Substitute $s=\frac35$, $c=\frac25$:

First term: $\frac{27c^4}{16s^3}=\frac{27\left(\frac25\right)^4}{16\left(\frac35\right)^3}$ $=\frac{27\cdot\frac{16}{625}}{16\cdot\frac{27}{125}}$ $=\frac{\frac{432}{625}}{\frac{432}{125}}=\frac{125}{625}=\frac15$

Second term: $\frac{c}{2}=\frac{1}{2}\cdot\frac25=\frac15$

Therefore, $\frac{27\csc^6\theta+8\sec^6\theta}{16\sec^8\theta}=\frac15+\frac15=\frac25$

6. Comparing with the options, the correct answer is $\boxed{\frac{2}{5}}$ which is option A.