jmc

Since $OY$ is a radius of the circle with centre $O,$ we have $OY=12.$ To find the length of $XY,$ we must find the length of $OX.$

Since $OA=OB,$ we know that $△OAB$ is isosceles.

Since $\angle AOB=60^{\circ },$ we have $$\angle OAB=\frac{1}{2}(180^{\circ }-60^{\circ })=60^{\circ }.$$ Therefore, $$\angle AOX=180^{\circ }-60^{\circ }-90^{\circ }=30^{\circ },$$ so $△OAX$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle.

Since $OA=12,$ we have $AX=\frac{1}{2}OA=6$ and $OX=\sqrt{3}AX=6\sqrt{3}.$ Thus, $$XY=OY-OX=\boxed{12-6\sqrt{3}}\approx \mathrm{1.61.}$$