You might want to give us the exact statement of the author.
Let the Wiener process $W_{s}$ be a r.v. from $\left(\mathcal{F}_{s},\Omega\right)\to\left(\mathcal{B}\left(\mathbb{R}\right),\mathbb{R}\right)$. The Borel-$\sigma$-algebra $\mathcal{B}\left(\mathbb{R}\right)$ contains all intervals of the form $\left[x,y\right]$ for $x\neq y\in\mathbb{R}$, because you have to be able to tell at time $s\geq 0$ if the Wiener process $W_{s}$ has its value in this interval or not. In order for $W_{s}$ to be measurable all the pre-images of this intervals have to be in the $\sigma$-algebra $\mathcal{F}_{s}^{W}$. So the (deterministic) random variable $X\left(t,\omega\right)=t$ is also measurable at time $s\geq 0$ because we can say in which interval its value is. But the deterministic r.v. $X\left(t,\omega\right)=t$ does not depend on $\omega$, so the pre-image of every obtainable resp. not obtainable interval is $\Omega$ resp. $\emptyset$.
Every deterministic r.v. is measurable to the trivial $\sigma$-algebra $\mathcal{F}_{0}:=\left\{\emptyset,\Omega\right\}$, which is contained in every other $\sigma$-algebra $\mathcal{F}_{s}^{W}$. So even if we condition on the coarser (smaller) $\sigma$-algebra $\mathcal{F}_{s}^{W}$ a deterministic r.v. is measurable and we only need the trivial $\sigma$-algebra $\mathcal{F}_{0}$. But that is
\begin{equation}
\mathbb{E}\left[t\mid\mathcal{F}_{0}\right] = \mathbb{E}\left[t\right] = t
\mathrm{.}
\end{equation}