The fraction is 
>!$\frac{2}{3}$


Without loss of generality let the radius of the smaller circle equal 1.[![enter image description here][1]][1]


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Edit:

Bubbler has observed in a comment that from here it is faster to continue like this
[![enter image description here][2]][2]

We have $JA=JD=r$, and since JH bisects the chord AB, $\angle{JHA}=90^\circ$.

We also have 

$$JA^2=HA^2+(HI^2+IJ^2)$$
$$r^2=1^2+(1^2+(1+\sqrt{3}-r)^2)$$
Solving for r,
$$r^2=1^2+1^2+r^2-2 \sqrt{3} r-2 r+2 \sqrt{3}+4$$
$$r=\frac{6+2\sqrt{3}}{2+2\sqrt{3}}=\sqrt{3}$$


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Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we have[![enter image description here][3]][3]


By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$[![enter image description here][4]][4]


Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.[![enter image description here][5]][5]
By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule
$$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$
$$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$
Combining these 2 equations,
$$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$
Solving for r produces
$$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$
We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$
which now gives 
$$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$
Squaring both sides,

>!$$r^2=\left(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}\right)^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$
>!$$r^2=3$$
>!$$r=\sqrt{3}$$

And the ratio of the areas is
>!$$\frac{\ \frac{2\cdot \pi \cdot (1^2)}{2}\ }{\ \frac{\pi \cdot (\sqrt{3})^2}{2}\ }=\frac{2}{3}$$


  [1]: https://i.sstatic.net/YvM5A.png
  [2]: https://i.sstatic.net/NEmls.png
  [3]: https://i.sstatic.net/LOvOa.png
  [4]: https://i.sstatic.net/xP2rV.png
  [5]: https://i.sstatic.net/IM9sE.png