So... after much discussions in the comments, in the end it seems that the letter $\mathcal N$ is used here to denote $\Phi$ the CDF of the standard normal distribution, defined by $\Phi(x)=\mathrm P(X\leqslant x)$, where $X$ is any standard normal random variable.
Since $W_t=\sqrt{t}Y$ where $Y$ is standard normal, $\mathrm E(\Phi(W_t))=\mathrm P(X\leqslant\sqrt{t}Y)=\mathrm P(Z\leqslant0)$ with $Z=X-\sqrt{t}Y$. Since $X$ and $Y$ are independent and centered normal, $Z$ is centered normal hence $Z$ is symmetric and $\mathrm E(\Phi(W_t))=\Phi(0)=\frac12$ for every $t$.
By the same decomposition, $\mathrm E(\Phi(W_t+a))=\mathrm P(Z\leqslant a)$. Since $Z$ is centered normal with variance $1+t$, $\mathrm P(Z\leqslant a)=\mathrm P(\sqrt{1+t}\cdot X\leqslant a)$, hence, for every $a$ and every nonnegative $t$,
$\color{red}{\mathrm E(\Phi(W_t+a))=\Phi\left(\frac{a}{\sqrt{1+t}}\right)}$
Exercise: Use this approach to completely and readily solve this nearly duplicate question of yours.