Assume that the displacement (s) of air is proportional to the pressure difference $$(\Delta p)$$ created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air ($$\rho$$) and the frequency (f). If $$\Delta p \sim 10\,\text{Pa}$$, $$v \sim 300\,\text{m/s}$$, $$\rho \sim 1\,\text{kg/m}^3$$, $$f \sim 1000\,\text{Hz}$$, then $$s$$ will be of the order of (take the multiplicative constant to be 1):

We recall the standard relation for a plane progressive sound wave that connects the pressure amplitude with the displacement amplitude:

$$\Delta p \;=\; \rho\,v\,\omega\,s,$$

where $$\rho$$ is the density of air, $$v$$ is the speed of sound, $$\omega$$ is the angular frequency, and $$s$$ is the displacement amplitude. The angular frequency is linked to the ordinary frequency by the well-known identity $$\omega = 2\pi f.$$

The problem statement tells us to “take the multiplicative constant to be 1,” which means we may drop the factor $$2\pi$$. Accordingly we approximate

$$\omega \approx f,$$

and the above formula simplifies to

$$\Delta p = \rho\,v\,f\,s.$$

We now isolate $$s$$:

$$s = \frac{\Delta p}{\rho\,v\,f}.$$

All the required quantities are supplied:

$$\Delta p = 10\;\text{Pa}, \quad \rho = 1\;\text{kg/m}^3, \quad v = 300\;\text{m/s}, \quad f = 1000\;\text{Hz}.$$

Substituting each value into the expression for $$s$$ we have

$$s = \frac{10}{\bigl(1\bigr)\bigl(300\bigr)\bigl(1000\bigr)} = \frac{10}{300\,000} = \frac{1}{30\,000}\;\text{m}.$$

To render this in more familiar units, we convert metres to millimetres. Using $$1\;\text{m} = 1000\;\text{mm},$$ we get

$$s = \frac{1}{30\,000}\;\text{m}\;\times\;1000\;\frac{\text{mm}}{\text{m}} = \frac{1000}{30\,000}\;\text{mm} = \frac{1}{30}\;\text{mm} \approx 0.033\;\text{mm}.$$

The numerical value $$0.033\;\text{mm}$$ can also be written as $$\dfrac{3}{100}\,\text{mm}.$$ Among the given options, this precisely matches option A.

Hence, the correct answer is Option A.