Answer
See the detailed answer below.
Work Step by Step
a) We know that the net torque, which is the straightening torque in our case, is given by
$$\sum\tau=I\alpha\tag 1$$
So we have to find the moment of inertia of the system and its angular acceleration.
The angular acceleration is constant, as the author told us, so it could be found by applying the kinematic formula of
$$\theta_f=\theta_i+\omega_i t+\frac{1}{2}\alpha t^2$$
where $\theta_i=0$, and $\omega_i=0$
$$\theta_f=0+0+\frac{1}{2}\alpha t^2$$
$$\alpha=\dfrac{2\theta_f}{t^2}\tag 2$$
Now the moment of inertia of the system which is the stamen plus the sac is given by
$$I=I_{stamen}+I_{sac}$$
where the stamen is similar to a uniform rod around its end.
$$I=\frac{1}{3}m_{\rm stamen}L_{\rm stamen}^2+m_{\rm sac}L_{\rm stamen}^2 $$
We can see that the mass of the sac is equal to the mass of the stamen, so
$$I=\frac{1}{3}m L^2+mL^2=\frac{4}{3}m L^2\tag 3$$
Plug (2) and (3) into (1) to find the straightening torque;
$$\sum\tau= \left[\frac{4}{3}m L^2\right]\left[\dfrac{2\theta_f}{t^2} \right]$$
Plugging the known;
$$\sum\tau= \left[\frac{4}{3}\times (10\times 10^{-6}\times 10^{-3})\times (1\times 10^{-3})^2 \right]\left[\dfrac{2\times 60^\circ\times \dfrac{2\pi}{360^\circ}}{(0.3\times 10^{-3})^2} \right]$$
$$\sum\tau=\color{red}{\bf3.1 ×10^{-7}}\;\rm N\cdot m $$
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b)
The final angular speed is given by
$$\omega_f=\omega_i+\alpha t$$
$$\omega_f=0+\alpha t$$
Plugging from (2);
$$\omega_f=\dfrac{2\theta_f}{t^{\color{red}{\bf\not} 2}} \color{red}{\bf\not} t=\dfrac{2\theta_f}{t}$$
Thus, the final speed is given by
$$v_f=\omega_f R=\omega_fL$$
where the radius here is the length of the stamen arm.
Hence,
$$v_f=\dfrac{2\theta_f}{t}L$$
Plugging the known;
$$v_f=\dfrac{2\times 60^\circ\times \dfrac{2\pi}{360^\circ}}{0.3\times 10^{-3}}\times 1\times 10^{-3}=6.98\;\rm m/s$$
$$v_f\approx \color{red}{\bf7.0}\;\rm m/s$$
