Answer
$v_f=34m/s$
Work Step by Step
Take the distance from one module to the center of mass (also where the axis of rotation passes) before they pull together to be $R$
The initial angular speed of each module is $\omega_0=\frac{17}{R}$
The initial moment of inertia is $I_0=\sum MR^2$
After they pull together, the distance $R$ becomes $R/2$. So, the final angular speed $\omega_f=\frac{v_f}{R/2}=\frac{2v_f}{R}$ and final moment of inertia $I_f=\sum \frac{MR^2}{4}$
Because angular momentum is conserved, $$\sum MR^2\times\frac{17}{R}=\sum \frac{MR^2}{4}\times\frac{2v_f}{R}$$ $$17=\frac{1}{4}\times2v_f$$ $$\frac{v_f}{2}=17$$ $$v_f=34m/s$$
