The square spinning about the center can be viewed as four smaller squares (half the size) spinning about the corner. To every surface element $dx\:dy$ at distance $r$ from the axis on the second square belongs therefore a surface element $dx\:dy$ at distance $\tfrac12r$ on the first square.

The torque on each square is the integral of $\eta\sigma gr$ over the surface. Therefore the torque on the second square is twice as much as on the first square.

The rotational inertia is the surface integral of $\sigma r^2$ . Therefore the rotational inertia of the second square is four times larger than that of the first square.

The rotational acceleration $\alpha = \tau/I$ of the second square is therefore $4/2 = \tfrac12$ of that of the first square.

Finally, since $t = \omega_0/\alpha$ , the time needed to slow the second square is twice as much as for the first square.

We can use scaling arguments to compare the torque on the two squares. Divide the first square into 4 small squares having sides $\dfrac{\ell}{2}$ and meeting at center.

Since the total torque about identical configuration (the corners in this case) for a small square and a larger one will scale in proportion to its mass and length, $\left (\tau= \displaystyle \int \mu g r~dm \right )$ so $\tau_2=\dfrac{4m}{m} \dfrac{ \ell}{\ell/2} \tau_0$ and also $\tau_1=4 \tau_0$ .

Now as for the moment of inertia it varies in proportion with $mr^2$ (we can use the nice symmetries of regular objects in using scaling arguments for $I$ ) and thus $I_2=2^2 2^2 I_0=4 I_1$ .

Using $\tau_{\text{axis}}= I_{\text{axis}} \alpha$ , we get $\alpha_2=\dfrac{\alpha_1}{2}$ .

The time taken to stop $t =\dfrac{\Delta \omega}{ \alpha}$ is $t_2=2 t_1.$

Nishant Rai plz post a solution