Consider a Poisson process with rate $\lambda$. Since the Inter-arrival time follows exponential distribution, the expectation value of the inter-arrival time in general is $E[X]= \frac 1 \lambda$.
If I consider a particular time interval $[0,1 \text{ s}]$, Then, what would be the expectation value of the arrival time $X_1,X_2,X_3,\ldots$ of $1^\text{st}, 2^\text{nd}, and 3^\text{rd}$ arrivals,......? provided that these arrival time values lie within $[0,1 \text{ s}]$.
My Question is, if I know the expectation value of the arrival time of the $1^\text{st}$ arrival, as $E[X_1\mid 0<X_1<1\text{ s}]$.
Its obvious that the expected value of the arrival time of the $2^\text{nd}$ would be greater than the $1^\text{st}$ arrival. Therefore, is it right to compute, $$E[\ X_2\mid (E[X_1\mid 0<X_1<1\text{ s}]) <X_2< 1\text{ s}].$$
You ask for $E(X_k|X_k<1)$ where $X_k$ is the arrival time of the $k$-th arrival in the Poisson process, which is distributed gamma($k,\lambda$) (note the condition $X_k>0$ is redundant). This is, by definition, equal to $$E(X_k \cdot 1_{X_k<1})/P(X_k<1),$$ where $1_A$ represents the indicator function of the set $A$.
The PDF of $X_k$ is $f_{X_k}(x)=\frac{\lambda^k}{(k-1)!} x^{k-1} e^{-\lambda x}$ and $P(X_k<1)=F_{X_k}(1)$ where $F$ is the CDF. Thus, the above expression can be written as, $$\frac{\int_0^1 \frac{\lambda^k}{(k-1)!} x^{k} e^{-\lambda x} dx}{1-e^{-\lambda}(1+\lambda+\lambda^2+\dotso+\lambda^{k-1})}$$
By the way, the whole idea of conditioning on the event that the mean arrival time of the first arrival is less than the mean time for the second arrival is moot. Expected values are no longer random, so the probability $P(E(X_1)<E(X_2))$ is undefined hence so is the conditional expectation $E(X_2 | E(X_1)<E(X_2))$. Any questions?
EDIT For arbitrary $t>0$, we have $$E(X_k | X_k<t)=\frac{\int_0^t \frac{\lambda^k}{(k-1)!} x^{k} e^{-\lambda x} dx}{1-e^{-\lambda t}(1+\lambda t+(\lambda t)^2+\dotso+(\lambda t)^{k-1})}$$
and specifically for $k=1$ and $k=2$, i.e. the first and second arrival times, this is simplified to, $$E(X_1 |X_1 <t)=\frac{1}{\lambda} +\frac{t}{1-e^{\lambda t}}$$ and
$$E(X_2 | X_2<t)=\frac{2}{\lambda} +\frac{\lambda t^2}{\lambda t+1-e^{\lambda t}},$$ although you should check these yourself by computing the expression for the values of $k$.