= Solution
Using <rotational kinetic energy> $K=I\omega^2/2$ at the two steady endpoint configurations,
$$
\boxed{K_i=155.150\,\mathrm J,\qquad K_f=539.593\,\mathrm J,\qquad
\Delta K=384.443\,\mathrm J.}
$$
At fixed <angular momentum> $J$, $K=J^2/(2I)$ increases when the <moment of inertia> decreases. The skater supplies <work> by pulling the arms inward; chemical energy in the muscles is converted into additional <rotational kinetic energy>, with possible extra dissipation in a real skater. This does not require an external axial <torque>.
For the ideal rotating <point masses>, an inward radial <force> supplies the required centripetal acceleration. During contraction both that inward <force> and the radial displacement point inward, so their scalar product gives positive <work>. At the initial and final steady configurations radial motion has ceased; any transient radial <kinetic energy> must also be included in a full intermediate-time balance. The quoted difference compares the rotational endpoint energies only.
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