The equation is certainly correct for measuring the moment of inertia of the object about the vertical going through the C.M.
1) Let L be the string length, D the distance between the two attached strings, W the weight of the object.
2) Rotate the object slightly in the plane.
In the following, the rotation is assumed so small that all cosines are approximated by unity.
There are two triangles two consider on each side:
a) The triangle in the vertical with the string length as the hypotenuse, and the displacement vector in the horizontal plane (normal to the direction given in that plane by the positions of the attachment points of strings in the undisplaced state).
This displacement vector has length [tex]L\sin\phi[/tex]
Clearly, the component of string tension relevant for rotation in the horizontal plane, is
[tex]\frac{W}{2}\sin\phi[/tex] for a single string.
b) The triangle in the horizontal plane with D/2 as the hypotenuse; clearly we have, for the displacement vector:
[tex]\frac{D}{2}\sin\theta={L}\sin\phi\to\sin\phi=\frac{D}{2L}\sin\theta[/tex]
Here, [tex]\theta[/tex] is the displacement angle in the plane.
c)Hence, the torque from one string is [tex]\frac{D^{2}W}{8L}\sin\theta[/tex]
d)The angular frequency fulfills therefore the relation:
[tex]\omega^{2}= \frac{D^{2}W}{4LI}[/tex]
where I is the moment of inertia of the object.
The given equation is a simple rewriting of that equation.