The angular momentum of the two masses is computed independent of the skater - you were given the total angular momentum of the skater (including arms and hands which are normally considered part of the person) and ONLY have to compute the moment of inertia / angular momentum of the masses. A point mass at the end of a string has $$I=mr^2$$ as you know. The arms of the skater were already accounted for, and the mass of the weights is not distributed along the arms, it is all at the end.
Angular momentum is $I\omega$. You should now be able to compute it from $I_{total}=I_{skater}+I_{masses}$, and $\omega$ is given. It will, of course, not change when the skater pulls in her arms - conservation of angular momentum, and there is no external torque on the skater-plus-masses system.
The moment of inertia of the masses does change when the skater pulls in her arms - you can compute it for the masses, but not for the arms (which are also coming closer). That is a problem with the question - you must assume a massless arm if you want to compute the moment of inertia when the arms are pulled in.
And you need the moment of inertia for the last part, since you can write the angular kinetic energy as
$$KE = \frac12 I \omega^2$$
So it is not enough to know $L$, you actually need to be able to compute the new angular velocity. And for that you must make a simplifying assumption (massless arms).
On that assumption, you can compute the increased kinetic energy from the above (because you know the new angular velocity from the new moment of inertia).
Before:
$$\omega = 10 rad/s\\
I_{skater}=50 kg m^2\\
I_{weights} = 10 kg m^2\\
KE = \frac12 I_{total} \omega^2$$
and you should be able to figure the rest from here...
Best Answer
If you included the negative sign in your answer, then you aren't (quite) answering the question that was asked. Magnitudes are positive by definition; the negative sign on an acceleration (or velocity) is simply there to indicate direction.