= Solution
For a Poisson random variable of mean $\lambda$,
$$
\langle n\rangle=\lambda,
\qquad
\langle n(n-1)\rangle=\lambda^2.
$$
Substitution into part (b) gives the closed Riccati equation
$$
\boxed{\dot\lambda=\alpha+\beta\lambda-\gamma\lambda^2.}
$$
Its positive equilibrium is stable on the nonnegative half-line and equals
$$
\lambda_*=\frac{\beta+\sqrt{\beta^2+4\alpha\gamma}}{2\gamma}.
$$
Thus under the Poisson approximation,
$$
\boxed{\lim_{t\to\infty}\langle n(t)\rangle
=\frac{\beta+\sqrt{\beta^2+4\alpha\gamma}}{2\gamma}.}
$$
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