Do enunciado, temos que $n = \sin^2 \alpha \cos^2 \alpha = \large{\frac{1}{25}}$ . Simplifiquemos agora a expressão:$$\frac{\sin^4\alpha + \cos^4\alpha}{\sin^6\alpha + \cos^6\alpha} = \frac{\overbrace{(\sin^2\alpha + \cos^2\alpha)^2}^{1} - 2n}{\underbrace{(\sin^2\alpha + \cos^2\alpha)}_{1}[\underbrace{(\sin^2\alpha + \cos^2\alpha)^2}_{1} - 3n]}$$Assim:$$\frac{\sin^4\alpha + \cos^4\alpha}{\sin^6\alpha + \cos^6\alpha} = \frac{1 - 2n}{1 - 3n} = \frac{1 - \frac{2}{25}}{1 - \frac{3}{25}} = \boxed{\frac{23}{22}}$$ $$\text{Alternativa } \mathbb{(B)}$$