Answer
$5.8\;\rm rev/s$
Work Step by Step
We just need to sketch the two situations, as seen below so that we can find the final angular velocity of the skier.
We plugged all the data in the figures below.
$\bullet$$\bullet$ In the first case, the skater is modeled as a cylindrical torso (160 cm long) plus two rod-like arms attached to the outside of the torso. So her angular momentum is given by
$$L_i=I_i\omega_i $$
we know that she was spinning at an angular speed of 1 rev/s. Hence,
$$L_i=I_i\tag 1$$
where the initial moment of inertia is given by $$I_i=I_{torso}+2I_{rods}$$
$$I_i=\frac{1}{2}M_{torso}R_{torso}^2+2I_{rods}\tag 2$$
The rods are rotating 10 cm away from their ends, as seen in the first figure below. So we can find their moment of inertia by using the parallel-axis theorem.
$$I_{rod}=\frac{1}{12}M_{rod}L_{rod}^2+M_{rod}d^2$$
where $L$ here is the length of the rod which represents the arm of the skater.
Noting that $d=\frac{1}{2}L_{rod}+R_{torso}$
$$I_{rod}=\frac{1}{12}M_{rod}L_{rod}^2+M_{rod}(\frac{1}{2}L_{rod}+R_{torso})^2 $$
Plugging into (2);
$$I_i=\frac{1}{2}M_{torso}R_{torso}^2+2\left[\frac{1}{12}M_{rod}L_{rod}^2+M_{rod}(\frac{1}{2}L_{rod}+R_{torso})^2\right]$$
Plugging into (1) and then plugging the given to find her initial moment of inertia
$$L_i=\frac{1}{2}M_{torso}R_{torso}^2+ \frac{1}{6}M_{rod}L_{rod}^2+2M_{rod}(\frac{1}{2}L_{rod}+R_{torso})^2 $$
$$L_i=\frac{1}{2}(40)(0.10)^2+ \frac{1}{6}(2.5)(0.66)^2+2(2.5)(0.33+0.10)^2 $$
$$L_i=\bf 1.306\;\rm kg\cdot m^2\cdot rev/s\tag 3$$
$\bullet$$\bullet$ In the second case, the skater is modeled as a cylindrical torso(200 cm long).
$$L_f=I_f\omega_f $$
where the final moment of inertia is given by $$I_f=I_{torso}=\frac{1}{2}M_{torso}R_{torso}^2$$
Thus,
$$L_f=I_{torso}=\frac{1}{2}M_{torso}R_{torso}^2\omega_f =\frac{1}{2}(45)(0.10)^2\omega_f $$
$$L_f=0.225 \omega_f \tag 5$$
We assume that in both cases the skater is spinning without friction and then her angular momentum is conserved.
Thus,
$$L_i=L_f$$
Plugging from (3) and (5);
$$1.306=0.225\omega_f$$
Hence,
$$\omega_f=\dfrac{1.306}{0.225}=\color{red}{\bf 5.8}\;\rm rev/s$$
