= Solution
Use the convention $J_0=0$, so $A(t)=t$ when $N_t=0$. Conditional on $N_t=n\geq1$, the last jump $J_n$ is the largest of $n$ independent uniform points on $[0,t]$. The stated <order statistic> formula gives
$$
\mathbb E(J_n\mid N_t=n)=\frac{nt}{n+1},
$$
and hence, for every $n\geq0$,
$$
\mathbb E(A(t)\mid N_t=n)=\frac{t}{n+1}.
$$
Putting $\mu=\lambda t$ and averaging over $N_t$,
$$
\begin{aligned}
\mathbb EA(t)
&=t e^{-\mu}\sum_{n=0}^\infty
\frac{\mu^n}{n!(n+1)}\\
&=t e^{-\mu}\frac{e^\mu-1}{\mu}\\
&=\boxed{\frac{1-e^{-\lambda t}}{\lambda}}.
\end{aligned}
$$
This agrees with <age of a Poisson process>.
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