= Solution
The equilibrium-delay theorem for a <renewal process> with positive holding time $S$, finite mean $\mu$, and distribution function $F$ says that the first future renewal must have the <equilibrium residual-life distribution>,
$$
F_e(t)=\frac1\mu\int_0^t(1-F(s))\,ds\quad(t\ge0),
$$
independently of subsequent ordinary holding times. This is the residual lifetime seen at a stationary observation time, rather than an ordinary inter-renewal lifetime.
Here $\mu=3/\lambda$ and $1-F(t)=e^{-\lambda t}(1+\lambda t+\lambda^2t^2/2)$. Thus the required initial-delay density is
$$
\boxed{f_{S_1^D}(t)=\frac\lambda3e^{-\lambda t}\left(1+\lambda t+\frac{\lambda^2t^2}{2}\right)\mathbf1_{t>0}.}
$$
Equivalently, choose with equal probability a sum of one, two, or three independent rate-$\lambda$ <exponential random variables>. The three <Erlang distributions> have densities that average to the displayed density. A concrete stationary construction chooses the phase modulo three of a stationary Poisson clock uniformly; the number of exponential stages until its next third-stage renewal is then uniform on $\{1,2,3\}$.
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