The apparatus in the diagram consists of a solid cylinder of radius $\text{1}$ cm attached at the center to two disks of radius $\text{2}$ cm. It is placed on a surface where it can roll, but will not slip. A thread is wound around the central cylinder. When the thread is pulled at the angle $\theta = \text{90}^{\circ}$ to the horizontal (directly up), the apparatus rolls to the right. Which below is the largest value of $\theta$ for which it will not roll to the right when pulling on the thread?
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[/asy]
(A) $\theta = \text{15}^{\circ}$
(B) $\theta = \text{30}^{\circ}$
(C) $\theta = \text{45}^{\circ}$
(D) $\theta = \text{60}^{\circ}$
(E) None, the apparatus will always roll to the right
To determine the largest value of $\theta$ for which the apparatus will not roll to the right when pulling on the thread, we need to analyze the forces and torques acting on the system.

1. **Identify the forces and torques:**
   - The force \( F \) is applied at an angle \( \theta \) to the horizontal.
   - The radius of the central cylinder is \( r = 1 \) cm.
   - The radius of the disks is \( R = 2 \) cm.

2. **Condition for rolling without slipping:**
   - For the apparatus to roll without slipping, the torque due to the applied force \( F \) must be balanced by the torque due to the frictional force at the point of contact with the surface.

3. **Calculate the torque due to the applied force:**
   - The torque \( \tau \) due to the applied force \( F \) is given by:
     \[
     \tau = F \cdot r \cdot \sin(\theta)
     \]
   - Here, \( r \) is the radius of the central cylinder, and \( \theta \) is the angle at which the force is applied.

4. **Calculate the torque due to the frictional force:**
   - The frictional force \( f \) acts at the point of contact with the surface, which is at a distance \( R \) from the center of the apparatus.
   - The torque due to the frictional force is:
     \[
     \tau_f = f \cdot R
     \]

5. **Balance the torques:**
   - For the apparatus to not roll to the right, the torque due to the applied force must be less than or equal to the torque due to the frictional force:
     \[
     F \cdot r \cdot \sin(\theta) \leq f \cdot R
     \]

6. **Determine the maximum angle \( \theta \):**
   - The maximum angle \( \theta \) occurs when the torques are equal:
     \[
     F \cdot r \cdot \sin(\theta) = f \cdot R
     \]
   - Solving for \( \theta \):
     \[
     \sin(\theta) = \frac{f \cdot R}{F \cdot r}
     \]
   - Given that \( f \leq F \) (since friction cannot exceed the applied force), the maximum value of \( \sin(\theta) \) is:
     \[
     \sin(\theta) = \frac{R}{r}
     \]
   - Substituting the given values \( R = 2 \) cm and \( r = 1 \) cm:
     \[
     \sin(\theta) = \frac{2}{1} = 2
     \]
   - However, since the sine of an angle cannot exceed 1, we need to reconsider the physical constraints. The correct approach is to use the ratio of the radii directly:
     \[
     \cos(\theta) = \frac{r}{R} = \frac{1}{2}
     \]
   - Therefore:
     \[
     \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ
     \]

The final answer is \( \boxed{ \theta = 60^\circ } \).