Fix $T>0$. Write $C$ for the space of all continuous functions $f:[0,T] \to \mathbb{R}$. For a standard Brownian motion $B = \{B_t\}_{t \in [0,T]}$ defined on probability space $(\Omega,F,P)$, let $W$ be the classical Wiener measure. That is, we treat $B$ as a random variable $B: \Omega \to C$ and $B(\omega)\in C$ is the continuous sample path, so $W$ is the distribution law of $B$: if $E \subseteq C$ is a measurable set, then $W(E) = P(\{\omega \in \Omega | B(\omega) \in E\})$.
Now fix $t \in [0, T]$. I am asked to compute this integral:
\begin{equation*} \int_{C}\phi(t)dW(\phi) \end{equation*}
How do I even start? I have no idea how to approach such a problem. In a special case which I integrate on a finite set $E=\{\phi_1,\cdots,\phi_n\} \subseteq C$, I can write:
\begin{equation*} \int_{E}\phi(t)dW(\phi) = \sum_{i=1}^n\phi_i(t)W(\{\phi_i\}) = \sum_{i=1}^n\phi_i(t) P\{B(\omega) = \phi_i\} \end{equation*}
Even in such a special case I don't know how to calculate $P\{B(\omega) = \phi_i\}$ .
This all may become simpler to grasp if you add some formalism. You have a measure $W$ on $C$ and you are asked to compute an integral $$ \int_C e_t(\phi)W(\mathrm d\phi) $$ where $e_t:C \to \Bbb R$ is the evalution map that assigns to each function $\phi\in C$ its value $\phi(t) \in \Bbb R$. So you can treat $(C,\mathcal C, W)$ as a probability space and hence $e_t$ would just be a random variable, whose expectation you are asked to find. You very well know the distribution of this random variable: $$ W(\phi:e_t(\phi) \in A) = P(\omega: B(t, \omega) \in A) = \mathcal N(0, t)(A). $$ The rest should be easy.