The correct answer is C. 300 mm.
The maximum bending stress in a simply supported beam is given by the following equation:
$\sigma_{max} = \frac{M}{S}$
where:
$\sigma_{max}$ is the maximum bending stress, $M$ is the bending moment, and $S$ is the section modulus.
The bending moment in a simply supported beam with a uniformly distributed load is given by the following equation:
$M = \frac{wl^2}{8}$
where:
$w$ is the load per unit length, $l$ is the span length, and $M$ is the bending moment.
The section modulus of a rectangular beam is given by the following equation:
$S = \frac{bh^2}{6}$
where:
$b$ is the width of the beam, $h$ is the depth of the beam, and $S$ is the section modulus.
Substituting the equations for $M$ and $S$ into the equation for $\sigma_{max}$, we get the following equation:
$\sigma_{max} = \frac{w\frac{l^2}{8}}{bh^2/6} = \frac{wl^2}{4bh}$
We are given that the load per unit length is $w = 75 \frac{N}{m}$, the span length is $l = 8 \text{ m}$, the maximum bending stress is $\sigma_{max} = 75 \frac{N}{mm^2}$, and the modulus of elasticity is $E = 0.2 \times 10^6 \frac{N}{mm^2}$. Substituting these values into the equation for $\sigma_{max}$, we get the following equation:
$75 \frac{N}{mm^2} = \frac{(75 \frac{N}{m})(8 \text{ m})^2}{4bh}$
Solving for $h$, we get the following equation:
$h = \frac{(75 \frac{N}{m})(8 \text{ m})^2}{4(75 \frac{N}{mm^2})(0.2 \times 10^6 \frac{N}{mm^2})} = 300 \text{ mm}$
Therefore, the depth of the joist is 300 mm.