$$E\left[\int_0^\infty \left|\int_0^t(W(s))^2 \, dW(s)\right|^{\,2} e^{-t} \, dt\right]$$
Not sure if, by Itô Isometry, it is possible to pull the expectation operation inside in the integral because of the $$e^{-t}$$ term.
Any insight is appreciated.
Let $X_t = \int_0^t W_s^2 dW_s$. You want to compute $C = \mathbb{E}[\int_0^\infty X_t^2 e^{-t} dt]$. Since $X_t^2 e^{-t} \geq 0$, by Fubini's Theorem we have \begin{align*} C =& \int_0^\infty \mathbb{E}[X_t^2 e^{-t}] dt \\ =& \int_0^\infty e^{-t} \mathbb{E}\left[\int_0^t W_s^4 ds\right] dt \\ =& \int_0^\infty e^{-t} \int_0^t \mathbb{E}[W_s^4] ds dt \\ =& \int_0^\infty \int_0^t 3e^{-t}s^3 ds dt \\ =& \int_0^\infty t^3 e^{-t} dt \end{align*} where the second equality follows from the Ito Isometry and the third by another application of Fubini.