1. Use single-slit diffraction condition

For a slit of width $a$, the first minima occur at

$a \sin \theta = \lambda$

where $\theta$ is the angle of the first minimum from the central axis.

2. Substitute the given values

Given:

$\lambda = 2.0\,\text{cm}, \qquad a = 4.0\,\text{cm}$

So,

$\sin \theta = \frac{\lambda}{a} = \frac{2.0}{4.0} = \frac{1}{2}$

Hence,

$\theta = 30^\circ$

3. Find angular spread of the central maximum

The central maximum extends from the first minimum on one side to the first minimum on the other side.

Therefore, angular spread is

$2\theta = 2 \times 30^\circ = 60^\circ$

4. Match with the options

Thus, the correct option is:

$\boxed{60^\circ}$

So, Option A is correct.