1. Torque due to the applied force

The force is applied tangentially at radius $r=2\,\text{m}$, so the torque is

$\tau(t)=rF=2(20t-5t^2)=40t-10t^2$

2. Angular acceleration

Using $\tau=I\alpha$ with $I=10\,\text{kg m}^2$,

$\alpha(t)=\frac{\tau}{I}=\frac{40t-10t^2}{10}=4t-t^2$

3. Angular velocity

Assuming the pulley starts from rest,

$\omega(t)=\int \alpha(t)\,dt=\int (4t-t^2)dt=2t^2-\frac{t^3}{3}$

(using $\omega(0)=0$)

The direction reverses when angular velocity again becomes zero:

$2t^2-\frac{t^3}{3}=0$ $t^2\left(2-\frac{t}{3}\right)=0$

So, apart from $t=0$,

$t=6\,\text{s}$

Thus the pulley rotates in the initial direction up to $t=6\,$s, and then reverses.

4. Angular displacement before reversal

Now,

$\theta = \int_0^6 \omega(t)\,dt = \int_0^6 \left(2t^2-\frac{t^3}{3}\right)dt$

$\theta = \left(\frac{2t^3}{3}-\frac{t^4}{12}\right)_0^6$

At $t=6$,

$\theta = \frac{2(216)}{3}-\frac{1296}{12} = 144-108=36\,\text{rad}$

5. Number of rotations

$N=\frac{\theta}{2\pi}=\frac{36}{2\pi}=\frac{18}{\pi}\approx 5.73$

So the pulley makes about $5.73$ rotations before reversing.

6. Option check

Since $5.73$ is more than $3$ but less than $6$, the correct option is:

$\boxed{\text{A}}$