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).
This is where it's a good time to "converse with the math". Let's look at the equation:
$$\mathcal{E} = \oint_\gamma \mathbf{E} \cdot d\mathbf{l}$$
which in this case equals $\mathcal{E}_\mathrm{ind}$. This is based on the work formula:
$$W = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$
Thus, what your question is, is essentially, asking "how can you have negative work". Looking at the above equation and recalling how a dot product behaves, there's only one way: $\mathbf{F}$ and the element of displacement $d\mathbf{r}$ must be aimed at cross purposes with each other. In the case of the integral for $\mathcal{E}_\mathrm{ind}$, the same goes only with $\mathbf{E}$ and $d\mathbf{l}$.
Hence, the answer to your question is: when $\mathbf{E}$ points opposite to $d\mathbf{l}$.
But what does that mean? Well, the key here is that we have to think a little more closely about the work formula. I believe what you are imagining it means is "the work done by the force as the force pulls the particle along with it". It actually is more general - in a work integral, the particle can be moved in any direction, including against the force. Of course, to make that motion happen in real life, you need to supply a source of contrary force, but that doesn't change the maths. This is why, say, in a more elementary example, you can talk of "negative work" done by the gravitational force when you lift an object off the floor.
(Why is it defined that way? Well, for one, because we often can't solve for the "real" trajectory the particle follows! If we stipulated that as a precondition, it would make work an extremely non-trivial concept!)
Where your mistake lies in, then, is in missing that. The displacement around the wire $d\mathbf{l}$ that we use to describe emf is not the displacement that necessarily occurs in reality to a real positive charge (after all, in many applications we aren't "really" dealing with positive charges anyways!).
Rather, it is a hypothetical one where we imagine that we grab a charge and move it around all the way through the circuit in a specific, fixed direction, and ask what the work - whether positive or negative - done by the electric force for that movement is. If we ask about the work done in the actual movements of charges, we will get a different answer, and yes, this one will always be positive, at least provided we don't get into looking at the situation in too-fine detail.
From an intuitive point of view, if you want to figure when the force is doing "positive" work and when it's doing "negative" work, imagine that you can feel the electric force tugging on the positive charge in your hand as you move it through the circuit. When you feel it helping you, i.e. the tug is with the motion of your hand, at that moment (i.e. that small increment $d\mathbf{l}$) the electric force is doing positive work, and you are doing negative work (to retard the charge, if you were to try and not naturally speed up your hand as you'd likely tend to, of course). When you feel it is fighting you, i.e. the tug is against the motion of your hand, the electric force is doing negative work, and you are doing positive work (to help it against the contrary pull). Total negative work, and hence negative emf, will be if, in moving the charge, you had to fight more often than flow.
Best Answer
You just have to write $emf = Ri$ and integrate over time since $i=dQ/dt$.