= Solution
Let the motor's constant <power> be $P>0$ and put $\omega=\dot\theta$. The braking coefficient in the proposed formula must describe a retarding <torque> $k\omega$. This <constant-power rotational acceleration> obeys the <energy balance>:
$$
\frac d{dt}\left(\frac I2\omega^2\right)=P-k\omega^2.
$$
Consequently the steady <angular velocity> is $\omega_0=\sqrt{P/k}$, and the ordinary rotational <frequency> is $\omega_0/(2\pi)$. Solving the linear equation for $\omega^2$ from rest gives
$$
\omega^2(t)=\omega_0^2\left(1-e^{-2kt/I}\right),\qquad
\boxed{t=-\frac I{2k}\ln\left(1-\frac{\omega^2}{\omega_0^2}\right).}
$$
\b[The printed time formula is missing a minus sign.] Its logarithm is negative for $0<\omega<\omega_0$, whereas elapsed time must be positive. The terminal speed is approached only as $t\to\infty$.
A constant-power motor starting exactly from rest is an idealization: its driving <torque> $P/\omega$ would diverge. The energy equation has a regular solution for $\omega^2$, although the initial angular acceleration is singular. If $k\omega$ were literally a tangential force, a lever arm would also be needed; the dimensions of the requested expression identify $k$ as a rotational damping coefficient.
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