This is a special case of stochastic Fubini's theorem for deterministic integrands:
Let $f : [0,t] \times [0,t] \to\mathbb{R}$ be measurable. Assume that $$ \int_0^t \left( \int_0^t |f(r,s)|^2 dr \right)^{1/2} ds <\infty.$$ Then the integrals $I_1= \int_0^t \left( \int_0^t f(r,s)ds \right) dW_r $ and $I_2 = \int_0^t \left( \int_0^t f(r,s)dW_r\right) ds$ are well-defined and $I_1=I_2$ holds almost surely.
(end of the theorem)
Reference: a special case of Theorem 4.18 in "G. Da Prato and J. Zabczyk. Stochastic equations in infinite dimensions, volume 44 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 1992".
Let's now set the question of integral interchanging aside and focus on whether the integrals are well-defined. I've noticed that the integrability condition ensures that $I_2$ is well-defined and $I_2 \in L^2(\Omega)$. However, I don't see why the integral $I_1$ should be well-defined. I should furthermore require that $$ \int_0^t \left( \int_0^t f(r,s)ds \right)^2 dr <\infty.$$
Is this condition automatically satisfied by the condition from the theorem? How can I see that $I_1$ exists? Is there something else a fail to notice?