How to model this problem to a differential equation? Consider a sexually reproducing population. Assume the gender ratio of the population is constant so the population size can be assumed to be $N $. The encounter rate is proportional to the square of population size so $N^2$. On encounter, we assume animals produce offspring at a rate that decreases linearly with population size $N $ from a population capacity $L $, as in the logistic growth model with per capita birth rate $\beta $
Use these assumptions to construct the growth rate in terms of $N$, $L$ and $\beta $
 A: I think the expected answer is $\beta N^2 (1 - N/L)$. Here is how to obtain that.
$$\left(\text{growth rate}\right) = \left(\text{encounter rate}\right) * \left(\text{birth rate}\right)$$
Since the (encounter rate) is given as $N^2$, we have the first term. Now the logistic growth looks like this $r(1-x/K)$, where $r$ is the intrinsic growth rate (or per capita growth rate - your $\beta$ basically), then you have your birth rate is $\beta (1 - N/L)$ in an analogous way. Thus you obtain your growth rate.

Well, here is the problem. Although no model is perfect, this way of constructing the growth term is questionable. If you notice, the solution is dimensionally incorrect for a growth rate. The main problem lies in the shortcut/assumption that the encounter rate is simply $N^2$.
One extra bit of information that makes this model slightly different than others is the constant ratio of male to female populations. Let's give it a name, say $\alpha$, and also call the male population $M$ and female population $F$, then $\frac{M(t)}{F(t)} = \alpha$, or $M = \alpha F$. Since $M + F = N$, then $\alpha F + F = N$, or the proportion of female in the total population is:
$$\frac{F(t)}{N(t)} = \frac{1}{1 + \alpha}$$
Assume homogeneous mixing of the population, which means the chance of encounter a female is the same everywhere, then the encounter rate is the product of (the total male population) times (the proportion of female in the total population), which means:
$$\left(\text{encounter rate}\right) = \frac{1}{1+\alpha}M(t)$$
Thus what I believe should be the correct answer is:
$$\frac{\beta}{1+\alpha}M(t)\left(1-\frac{N(t)}{L}\right)$$
Additionally, using $M = \alpha F = \alpha \frac{N}{1+\alpha}$, this becomes
$$\frac{\beta\alpha}{(1+\alpha)^2}N(t)\left(1-\frac{N(t)}{L}\right)$$
This has the correct dimension for a growth rate and also contains the ratio term $\alpha$, a given information that is not justly used (therefore it is not the answer that is expected).
