Correct Answer: $\frac{256}{481}$

Solution : Given: $\frac{\cos \beta}{\sec \alpha}=15$ and $\frac{\sin \beta}{\sec \alpha}=16$
$\frac{\cos \beta}{\sec \alpha}=15$
⇒ $\frac{\cos \beta}{15}=\sec \alpha$......................(equation 1)
Again, $\frac{\sin \beta}{\sec \alpha}=16$
⇒ $\frac{\sin \beta}{16}=\sec \alpha$........................(equation 2)
From equations 1 and 2, we get:
$\frac{\cos \beta}{15}=\frac{\sin \beta}{16}$
⇒ $\frac{\cos \beta}{\sin \beta}=\frac{15}{16}$
⇒ $\cot\beta = \frac{15}{16}$
Since $\cot\beta = \frac{\text{Base}}{\text{Perpendicular}}$
$\text{Hypotenuse}=\sqrt{\text{Perpendicular}^2+\text{Base}^2} = \sqrt{16^2+15^2}=\sqrt{481}$
$\therefore \sin ^2 \beta=(\frac{\text{Perpendicular}}{\text{Hypotenuse}})^2=(\frac{16}{\sqrt{481}})^2$
= $\frac{256}{481}$
Hence, the correct answer is $\frac{256}{481}$.