= Solution
Let $\mu=\langle X\rangle$. The <Markov jump-process generator> applied to $X$ gives the exact <moment equation> $\dot\mu=3\lambda-\beta\langle\sqrt X\rangle$. With the prescribed <moment closure>, $\langle\sqrt X\rangle\approx\sqrt\mu$, stationarity gives
$$
\boxed{\mu\approx\left(\frac{3\lambda}{\beta}\right)^2.}
$$
The production flux counts molecules, not reaction events: it is $v^+=3\lambda$, while the removal flux is $v^-=\beta\sqrt\mu=3\lambda$. The <reaction-rate elasticity> of removal relative to constant production is
$$
H=\frac{\partial\log v^-}{\partial\log\mu}-\frac{\partial\log v^+}{\partial\log\mu}=\frac12.
$$
The <mean molecular lifetime> is population divided by its stationary molecular removal flux. Therefore
$$
\boxed{H=\frac12,\qquad \tau=\frac{\mu}{\beta\sqrt\mu}=\frac{\mu}{3\lambda}\approx\frac{3\lambda}{\beta^2}.}
$$
The noise-relevant average chemical event size is the <molecular-flux-weighted reaction event size>. For the <stoichiometric vectors> $+3$ and $-1$ it is
$$
\langle s\rangle=\frac{3^2\lambda+(-1)^2\beta\sqrt\mu}{3\lambda+\beta\sqrt\mu}
=\frac{9\lambda+3\lambda}{3\lambda+3\lambda}
=\boxed{2.}
$$
This weights each absolute jump by the molecular turnover it contributes. Equivalently, the balanced production and removal fluxes give equal weight to their sizes $3$ and $1$, yielding $(3+1)/2$. An event-count-weighted average would be $3/2$, since three quarters of firing events are deaths at stationarity; that different average does not determine the diffusion term in the <linear noise approximation>.
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