Since, ${\sec }^2\alpha +{\mathrm{cosec}}^2\alpha -3=\left(1+{\tan }^2\alpha \right)+\left(1+{\cot }^2\alpha \right)-3$

$=-1+{\left(\tan \alpha -\cot \alpha \right)}^2+2=1+\left(\tan \alpha -\cot {\alpha }^2\right)>0$

Hence, the point $\left(\sec \alpha ,\cos ec\alpha \right)$ lies outside the circle $x^2+y^2-3=0$
Now, the minimum distance
$=\sqrt{{\left(\sec \alpha -0\right)}^2+{\left(\mathrm{cosec}\alpha -0\right)}^2}-\sqrt{3}$
$=\sqrt{4+{\left(\tan \alpha -\cot \alpha \right)}^2}-\sqrt{3}$ (Minimum is attained, when $\tan \alpha =\cot \alpha$)
$=2-\sqrt{3}\equiv a-\sqrt{b}$ (Given)
So, $a=2,\mathrm{}b=3.$

Hence, $\frac{b}{a}=\frac{3}{2}=1.5$