Answer
b) is correct.
Work Step by Step
We have $$KE_r=\frac{1}{2}I\omega^2=\frac{1}{2}L\omega$$
While the skater's angular momentum $L$ is conserved, her angular speed $\omega$ increases, so her rotational kinetic energy increases as she pulls her arms in.
Therefore, her final rotational kinetic energy is larger. b) is correct.
