I stumbled across this problem. I am not sure if I have approached it in a meaningful way. Any advice/help/correction would be much appreciated.
The following was my approach:
Any such chord would result in an isosceles triangle with lengths; 1,1 (both radii of unit circle) and x (where x= length of the chord).
Letting a= the angle between the radii, and $\frac {180-a}{2}$ = the remaining two angles, we obtain the following using the sine rule:
$\frac{x}{sin(a)}=\frac{1}{sin(\frac{180-a}{2})}$
And hence: $x=\frac{sin(a)}{sin(\frac{180-a}{2})}$
I then got the average value of this function between 0 and $\pi$ as follows:
$\frac{1}{\pi -0} \int_0^\pi \frac{sin(a)}{sin(\frac{\pi-a}{2})}$
And found it to be $\frac{4}{\pi}$ units.
As I say, I am unsure of my solution but found the question very interesting so any correction and/or confirmation of my work would be great.
It seems correct indeed we can also evaluate the mean length of a chord in a unit circle with respect to the point $(-1,0)$ with $\theta\in[-\pi/2,\pi/2]$ then
$$\frac2 \pi \int_0^{\pi/2} 2\cos \theta \,d\theta=\frac 4 \pi[\sin \theta]_0^{\pi/2}=\frac 4 \pi$$