The distance between two consecutive points with phase difference of $45^{\circ}$ in a wave of frequency 300…

To determine the wave velocity, first find the wavelength from the given phase difference and the separation distance between the two points. The phase difference $\Delta\phi$ and path difference $\Delta x$ are related by $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$.

The phase difference is $45^\circ$, equivalent to $\frac{\pi}{4}$ radians. With $\Delta x = 4.0$ m, substitute into the equation:
$\frac{\pi}{4} = \frac{2\pi}{\lambda} \times 4.0$

Divide both sides by $\pi$:
$\frac{1}{4} = \frac{8}{\lambda}$

Solving for $\lambda$ yields:
$\lambda = 32$ m

The wave velocity is given by $v = f \lambda$. With frequency $f = 300$ Hz:
$v = 300 \times 32 = 9600$ m/s

Convert to km/s:
$v = 9.6$ km/s

The correct option is $\boxed{\text{D}}$.