Let $\{X(t)\sim\text{Pois}(\lambda t)\}$ be a Poisson process with parameter $\lambda$, and $0<S_1<S_2<\dots$ be the occurrence time of the events $1,2,\dotsc$ and so on.
How do can find the pdf of $S_1|X(t)=n$ and use it to find $E[S_1|X(t)=n]$?
Also, what is the pdf of $S_i|X(t)=n$ for $i=1,2,\dotsc,n$? What would be $E[S_1+S_2+\dotsb+Sn|X(t)=n]$?
Hint: Notice that $P(S_1\in ds, X(t) = n)$ is the same as $$P(\text{First arrival in }ds\text{ and $n-1$ arrivals in (s,t]}).$$ Because these are disjoint blocks of time, the joint probability can be written as a product. Notice that $$P(S_1 \in ds) = \lambda e^{-\lambda s},$$ and that $X(t)-X(s)\sim \text{Pois}(\lambda(t-s)).$ Therefore $$P\{X(t)-X(s) = n-1\} = e^{-\lambda(t-s)}\frac{[\lambda (t-s)]^{n-1}}{(n-1)!}.$$ Putting it all together gives
\begin{align*} P(S_1\in ds|X(t) = n) &= \frac{P(S_1\in ds, X(t) = n)}{P(X(t) = n)}\\ &=\frac{\lambda e^{-\lambda s}e^{-\lambda(t-s)}[\lambda(t-s)]^{n-1}}{(n-1)!}\cdot \frac{n!}{e^{-\lambda t}(\lambda t)^n}\\ &=\frac{n(t-s)^{n-1}}{t^n}. \end{align*}
I think you can get the expectation from here. Also, apply this logic for the other $S_i$.