The correct answer is $\boxed{\text{B) 2.40 m}}$.
The volume of the ejector chamber is given by
$$V = \pi r^2 h$$
where $r$ is the radius of the chamber and $h$ is the height of the chamber. The peak sewage discharge is given by
$$Q = \frac{V}{t}$$
where $t$ is the time between discharges. Substituting the expression for $V$ into the expression for $Q$, we get
$$Q = \frac{\pi r^2 h}{t}$$
We are given that $h = 2 \text{ m}$ and $t = 10 \text{ min} = 600 \text{ s}$. We are also given that $Q = 0.0157 \text{ cumec} = 157 \text{ m}^3/\text{s}$. Substituting these values into the expression for $Q$, we get
$$157 \text{ m}^3/\text{s} = \frac{\pi r^2 (2 \text{ m})}{600 \text{ s}}$$
Solving for $r$, we get
$$r = \sqrt{\frac{157 \text{ m}^3/\text{s} \times 600 \text{ s}}{\pi (2 \text{ m})^2}} = 2.40 \text{ m}$$
Therefore, the diameter of the ejector chamber is $\boxed{\text{2.40 m}}$.
The other options are incorrect because they do not result in a diameter that is consistent with the given information.