We need to use the idea of conservation of angular momentum. This states, as applied to this particular example, that the skaters angular momentum will remain the same with her arms outstretched and with her arms above her head.
Let's let subscript 1 indicate the case where her arms are outstretched and subscript 2 indicate the case where her arms are over head.
We know that angular momentum is expressed as\[L = I \omega\]where \(I\) is the moment of inertia and \(\omega\) is the angular velocity in rad/s.
Refer here for a list of moments of inertia: http://en.wikipedia.org/wiki/List_of_moments_of_inertia
First, let's calculate the skaters moment of inertia in Case 1. Her moment of inertia can be represented as the sum of three separate, easy to represent geometric figures.
1) Her toros as a cylinder (8th row of the table at the above link.)
2 &3) Her arms as thin rods whose axis of rotation is at the end (3rd row)
This sum is expressed as\[I_1 = I_{Cyl} + I_{arm} + I_{arm} = I_{cyl} + 2 I_{arm} = {m_tr^2 \over 2} + 2 \cdot \left [m_aL^2 \over 3 \right ] \] where \(m_t\) is the mass of the skater's torso, \(r\) is the radius of the skaters torso, \(m_a\) is the mass of the skater's arms, and \(L\) is the length of the skater's arms.
The skater's moment of inertia in Case 2 can be modeled as a cylinder\[I_2 = {m_2 r^2 \over 2}\] where \(m_2 = 45 kg\) and \(r\) is the radius of the torso (same as in Case 1).
Now, we can set up an expression for the conservation of angular momentum as\[I_1 \omega_1 = I_2 \omega_2\]