= Solution
The normalized ansatz intended in the question is
$$
\phi(s,t)=\exp\bigl((s-1)f(t)\bigr),
$$
for which $\phi(1,t)=1$. Substitution into the equation from part (ii) and cancellation of $(s-1)\phi$ gives the <linear ordinary differential equation>
$$
f'=\lambda-\beta f.
$$
Thus
$$
f(t)=\frac\lambda\beta+\left(f(0)-\frac\lambda\beta\right)e^{-\beta t}.
$$
This generating function is that of a <Poisson distribution> with parameter $f(t)$. Equivalently, differentiating at $s=1$ gives
$$
\langle n\rangle=\partial_s\phi(1,t)=f(t),
\qquad
\sigma^2=\partial_s^2\phi(1,t)+\partial_s\phi(1,t)-\partial_s\phi(1,t)^2=f(t).
$$
Consequently
$$
\boxed{\langle n\rangle\longrightarrow\frac\lambda\beta,
\qquad \sigma^2\longrightarrow\frac\lambda\beta.}
$$
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