$Cov(X_t,X_s)$ of martingales 
Let  $X_t = \int_0^t W_u^2dW_u$ martingale
compute  : $$Cov(X_t,X_s)$$

note that $$Cov(\int_0^T a(t)dWt,\ \int_0^T b(t)dWt)\ = E[\int_0^T a(t)b(t)dWt]$$
My attempts:
$$Cov(X_t,X_s)\ = E[\int_0^{t\wedge s} W^{4}_udW_u]$$
Edit
if we use this :
$$\text{cov} \left( \int_0^T a(t) \, dW_t, \int_0^T b(t) \, dW_t \right) = \mathbb{E} \left( \int_0^T a(t) b(t) \, dt \right).$$
and choice $a(u) := 1_{[0,t]}(u) \cdot W_u^2,$ and $b(u) := 1_{[0,s]}(u) \cdot W_u^2,$
$T>\max\{t,s\}$
$$E[\int_{0}^T W_u^4 1_{u \leq t}1_{u\leq s}d_u] = E[\int_{0}^{s\wedge t} W_u^4 d_u]$$
In order to calculate the integral, we  use Fubini's theorem, i.e.
$$\mathbb{E}\left( \int_0^s W_u^4 \, du \right) = \int_0^{s\wedge t} \mathbb{E}(W_u^4) \, du.$$
to compute that $\mathbb{E}(W_u^4)=3u^2$ let's applying Ito's formula to $W_u^4$
$$dW_u^{4}=4W_u^{3}dW_u+\frac{1}{2}.4.3.W_u^{2}(dW_u)^2;\quad W_u^{4}(0))=0  $$
$$dW_u^{4}=4W_u^{3}dW_u+6W_u^{2}dt $$
Integrate:
$$W_u^{4}=4\int_{0}^{u}W_s^{3}dW_s+6\int_0^{u}W_s^{2}ds+0$$
$$\ldots $$
$$\mathbb{E}\left( \int_0^s W_u^4 \, du \right) = \int_0^{s\wedge t} \mathbb{E}(W_u^4) \, du.=\int_0^{s\wedge t} 3u^2 \, du=(s\wedge t)^3$$

The equation
$$\text{cov} \left( \int_0^T a(t) \, dW_t, \int_0^T b(t) \, dW_t \right) = \mathbb{E} \left( \int_0^T a(t) b(t) \, dW_t \right)$$
does not hold. Instead it should read
$$\text{cov} \left( \int_0^T a(t) \, dW_t, \int_0^T b(t) \, dW_t \right) = \mathbb{E} \left( \int_0^T a(t) b(t) \, \color{red}{dt} \right).$$
This follows from Itô's isometry and polarization.
