The moment of inertia of a rod of mass $m$ and length ($l$)about an axis passing through its centre and making an angle of $\theta$ w.r.t it is given as $\frac{m {l}^{2}}{12} {\sin}^{2} \theta$
here,as the axis passes through the bisector of the angle between the rods,so $\theta = 45$
So,w.r.t the mentioned axis, the rods have a total moment of inertia of $2 \cdot \frac{m {l}^{2}}{12} \cdot {\sin}^{2} 45 = \frac{m {l}^{2}}{12} K g . {m}^{2}$
Now,from perpendicular axis theorem, the moment of inertia of a ring of radius $\frac{l}{2}$ and mass $m$ w.r.t an axis passing through its centre and parallel to its plane is $\frac{m {\left(\frac{l}{2}\right)}^{2}}{2} = \frac{m {l}^{2}}{8} K g . {m}^{2}$
So,w.r.t the given axis,the moment of inertia of the whole system is $\frac{m {l}^{2}}{12} + \frac{m {l}^{2}}{8} = \frac{5 \left(m {l}^{2}\right)}{24} K g {m}^{2}$