$\mathbb{E}[(\sup_{1\leq t\leq\infty}W(t)/t)^2]$ where $W(t)$ is a Wiener process [closed]

I'm trying to find the value $\mathbb{E}[(\sup_{1\leq t\leq\infty}W(t)/t)^2]$. I know that $W(t)/t \sim N(0,1/t)$ and also that $\sup_{0\leq s \leq t}W(s) \overset{d}{=} |W(t)| $ but I don't see how to use that if we devide by $t$ and take the supremum over $t\in [1,\infty]$.

Let $W'(s):= s W(1/s)$ for $s>0$ and $W'(0)=0$. The time inversion property of the Brownian motion guarantees that the process $\left(W'(s)\right)_{s\geqslant 0}$ is a standard Brownian motion. As a consequence, $$ \left(\sup_{1\leqslant t<\infty}W(t)/t\right)^2=\left(\sup_{0< s\leqslant 1}sW(1/s)\right)^2=\left(\sup_{0< s\leqslant 1}W'(s)\right)^2\overset{\mathcal D}{=}\left(\sup_{0< s\leqslant 1}B(s)\right)^2 $$ where $B$ denotes a standard Brownian motion and by the recalled property, $$ \left(\sup_{1\leqslant t<\infty}W(t)/t\right)^2\overset{\mathcal D}{=}W(1)^2. $$