Answer
$1.2 kg \cdot m^2$
By bringing her arms close to her body.
Work Step by Step
There is no net external torque on the skater, so her net angular momentum is conserved.
$$L_{i}=L_{i}$$
$$I_i \omega_i = I_f \omega_f$$
$$I_f = \frac{\omega_i }{\omega_f}I_i $$
$$I_f = \frac{1.0rev/1.5s }{2.5rev/s}(4.6 kg \cdot m^2)=1.2 kg \cdot m^2$$
Physically, she pulls her arms closer to her body (the rotation axis) to reduce her moment of inertia.
