Variance of the Ito integral of W squared

Take a standard Brownian motion $W$ and form the stochastic integral $I_t=\int_0^t W_s^2\,dW_s$. Its variance has the shape $k\,t^3$. What is the constant $k$?

$k=1$. The integral is a mean-zero martingale, so by the Ito isometry its variance is $\int_0^t \mathbb{E}[W_s^4]\,ds=\int_0^t 3s^2\,ds=t^3$, using the normal fourth moment $\mathbb{E}[W_s^4]=3s^2$.