Assume an electron in a hydrogen atom undergoes a transition from n...

The angular momentum of an electron in the $n$-th orbit of a hydrogen atom is given by $L = \dfrac{nh}{2\pi}$.

The change in angular momentum during the transition from $n = 3$ to $n = 2$ is

$\Delta L = L_3 - L_2 = \dfrac{3h}{2\pi} - \dfrac{2h}{2\pi} = \dfrac{h}{2\pi}$

Substituting the value of Planck constant $h = 6.63 \times 10^{-34} \text{ J s}$, we get

$\Delta L = \dfrac{6.63 \times 10^{-34}}{2 \pi} \approx 1.05 \times 10^{-34} \text{ J s}$

The average torque acting on the electron is the rate of change of angular momentum:

$\tau = \dfrac{\Delta L}{\Delta t}$

Given the transition time $\Delta t = 10^{-8} \text{ s}$, the torque is

$\tau = \dfrac{1.05 \times 10^{-34}}{10^{-8}} = 1.05 \times 10^{-26} \text{ N m}$

The order of magnitude of the torque is $10^{-26} \text{ N m}$.