From $\int_0^1W_t\,dW_t={1\over 2}(W_t^2-t)$. One can deduce with the change of variable $s=t-\rho$, that your integral is equal to: $$ {1\over 2}(W_1^2-W_\rho^2-W_{1-\rho}^2) $$
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