Equilibrium will be reached when the net torque on the armature is zero. Since, as we will see below, it will be impossible to have the net torque vanish over an extended interval of time, we'll look for the situation when the torque averaged over time vanishes.
The Lorentz force law tells us that for a wire of length vector $\vec{l}$ carrying current $I$ in a uniform magnetic field $\vec{B}$, the force $\vec{F}$ is
\begin{equation} F = I \, \vec{l} \times \vec{B} \; \end{equation}
It follows that the strength of the torque $\tau$ on an armature with $N$ turns and cross-sectional area $A$ is given by
\begin{equation} \tau = N\, I \,A \,B \,\sin \theta \end{equation}
where $\theta$ is the direction between the normal vector of the armature cross-section and the magnetic field. (The geometry is actually slightly more involved than you might imagine at first - I recommend studying the diagram found in the bottom panel of this page).
We now want to know what happens if we take the time average of the torque over some time interval that spans many revolutions of the motor. $N$, $A$, and $B$ are all constant in this problem (although see the discussion at the very end of this answer concerning $B$), so the only quantities that can vanish in the average are $\langle I \rangle$ and $\langle \sin \theta \rangle$.
If this motor possesses no commutator, then the torque will switch directions every half cycle and it will subsequently be difficult for the armature to reach any appreciable speed at all. I will assume that the motor possesses a commutator that switches the current every half cycle. In this case, $\langle \sin \theta \rangle \not = 0 $ and so we must have $\langle I \rangle = 0$ at equilibrium.
In order to achieve $\langle I \rangle = 0$, it must be the case that $\langle \epsilon_\mathrm{net} \rangle = 0$ where $\epsilon_\mathrm{net}$ is the net emf and contains contributions from the 24 volt potential difference and the magnetic induction due to the rotation of the armature (but see the final note at the end of the answer for more details). As already pointed out in the statement of the question,
\begin{equation} \epsilon_\mathrm{net} = 24 \mathrm {\; V } - N \, A \, B \sin \theta \frac{d\theta}{dt}\end{equation}
and then taking the time average, for equilibrium to be achieved,
\begin{equation} \langle \sin \theta \frac{d\theta}{dt} \rangle = \frac{24 \mathrm {\; V }}{N \, A \, B}\end{equation}
Strictly speaking, $\langle \sin \theta \frac{d\theta}{dt} \rangle \not= \langle \sin \theta \rangle \langle\frac{d\theta}{dt} \rangle$ because $\frac{d\theta}{dt}$ will vary with time in a manner correlated to $\sin \theta$ (it is important to remember that we are actually averaging over time, not directly over $\theta$). A lower bound for $\frac{d\theta}{dt}$ is then the answer given in the back of the book. A more correct answer would be larger than that. In the approximation that $\frac{d\theta}{dt}$ is a constant in time (not actually true but perhaps quite a good approximation), then the answer would be multiplied by a factor of $\pi/2 \approx 1.57$.
In summary, I think the question was poorly constructed and you should not fret, in this case, about matching the book's answer exactly.
Final note concerning magnetic field from self-induction
There is an additional magnetic field generated from the current through the wires in the armature. As the current changes in time, the flux of this field through the armature will vary in time. This additional changing flux should in principle affect both the emf in the circuit and the torque on the armature.
However, these effects will be negligible if the magnetic field generated by the coils is negligible compared to the 0.33 Tesla magnetic field already in place. I urge you to calculate what is roughly the maximum current that could flow through the wires before the assumption that the corresponding field is negligible breaks down (I checked, and the current would have to become quite large indeed - but you should still check youself).
Best Answer
The magnetic flux as a function of time is $$ \Phi(t) = \mathbf B \cdot \mathbf A(t) = BA\cos(\omega t) $$ where $\mathbf B$ is the magnetic field and $\mathbf A(t)$ is the area vector as a function of time and $\omega t$ is the angle between the field and the area vector as a function of time. Then the rate of change of the flux as a function of time is $$ \Phi'(t) = -BA\omega\sin(\omega t) $$ Notice that if the angle between the area vector vector and the magnetic field is zero, then the flux is nonzero and equal to $AB$, its maximum, but the rate of change of the flux vanishes because $\sin(0) = 0$.