COM of a a Wire

Let's assume a co-ordinate system where $A$ is the origin and the angular bisector of $AB$ and $AC$ is the x-axis.

The length of $AB$ is $\dfrac{l}{2}$ and the location of its center of mass is the midpoint $(\frac{l}{4}\cos30^{\circ}, \frac{l}{4}\sin30^{\circ})$ .

Similarly, the center of mass of $AC$ is $(\frac{l}{4}\cos30^{\circ}, -\frac{l}{4}\sin30^{\circ})$ .

The centre of mass of the whole wire is the midpoint of the above two centers of mass (as the wire is of uniform mass distribution).

$\therefore C \equiv (\frac{l}{4}\cos30^{\circ}, 0)$

$\implies \text{required distance =} \frac{l}{4}\cos30^{\circ} = \boxed {\dfrac{l\sqrt3}{8}}$