The correct answer is C. ${{\text{m}}^{\frac{1}{3}}}$.
Lacey’s silt factor is a dimensionless number used in hydraulic engineering to estimate the silt load in a stream. It is defined as follows:
$$f = \frac{d_5}{D_84}$$
where $d_5$ is the median particle size of the silt in millimeters and $D_84$ is the 84th percentile particle size of the silt in millimeters.
The median particle size is the particle size that separates the silt into two equal parts, with half of the particles being smaller than $d_5$ and half of the particles being larger than $d_5$. The 84th percentile particle size is the particle size that separates the silt into two equal parts, with 84% of the particles being smaller than $D_84$ and 16% of the particles being larger than $D_84$.
Lacey’s silt factor is proportional to the median particle size of the silt to the power of $\frac{1}{3}$. This is because the median particle size is a measure of the average size of the silt particles, and the larger the average size of the silt particles, the more silt will be carried by the stream.
The other options are incorrect because they are not proportional to the median particle size of the silt to the power of $\frac{1}{3}$. Option A is proportional to the cube of the median particle size, which is not a meaningful measure of the silt load in a stream. Option B is proportional to the square root of the median particle size, which is a meaningful measure of the silt load in a stream, but it is not the correct power. Option D is proportional to the square of the median particle size, which is not a meaningful measure of the silt load in a stream.