I'm not sure if i completely understand the question, so please feel free to correct me.
Now, consider a torsional oscillator: f.ex. a mass at the end of a wire: The mass has the angular inertia, [itex]I[/itex], and the wire it's torsional constant, [itex]\tau_{c}[/itex], with values as specified. Assuming the total energy is conserved (no friction or outside forces), the total energy will be the sum of: (I) the potential energy, [itex]E_{pot}[/itex], corresponding to the work needed to rotate the mass to some angular displacement. (II) the kinetic energy, [itex]E_{kin}[/itex], due to the mass' velocity and angular inertia.
If you study the units of the torsional constant, [itex]\tau_{c}[/itex], you see that [N*m / rad] = [J / rad], applying that work is given from force times distance (Newton meters). Hence, if the energy you submitted is the total energy, [itex]E[/itex], then the maximum angular displacement, [itex]\theta_{max}[/itex], will be when,
[itex]
E_{kin}=0 \\<br />
E_{pot}= E = 5.4 J[/itex]
corresponding to the potential energy stored in the wire by 'twinning it up',
[itex]
\tau_{c} \cdot \theta_{max} = E_{pot}[/itex]
As you'll easily calculate yourself, inserting correct values and solving for [itex]\theta_{max}[/itex].
The maximum angular velocity, [itex]\omega[/itex], will then be at the point of rotation where,
[itex]
E_{pot}=0 \\<br />
E_{kin}= E = 5.4 J \\<br />
<br />
E_{kin} = \frac{1}{2} \cdot I \cdot \omega ^2[/itex]
Also there are all necessary values given: just solve for [itex]\omega[/itex]. That should be it. Good Luck!