Find the frequency response of the system
OK, in the comments you've shown that you can actually do it by yourself. This answer just summarizes the steps and shows you how to write your (correct) result in the form that you were given as an answer.

First of all, note that you have two identical systems in series. If $G(\omega)$ is the frequency response of the system up to and including the first integrator (from the left), then the total frequency response is

$$H(\omega)=G^2(\omega)\tag{1}$$

The response $G(\omega)$ can be seen as a multiplication of two subsystems, the system up to and including the adder, and the integrator. The first of the two has a frequency response

$$G_1(\omega)=1-e^{-j\omega T}\tag{2}$$

Since $G_1(\omega)$ has a zero at $\omega=0$, we don't need to bother with any problems of the integrator at $\omega=0$ (I'm talking about the delta impulse). So for calculating $G(\omega)=G_1(\omega)G_2(\omega)$, we can simply use

$$G_2=\frac{1}{j\omega}\tag{3}$$

which gives

$$G(\omega)=\frac{1-e^{-j\omega T}}{j\omega}\tag{4}$$

and, from (1),

$$H(\omega)=\frac{(1-e^{-j\omega T})^2}{(j\omega)^2}\tag{5}$$ 

In order to get from (5) the answer that you were given, use this trick to rewrite $G_1(\omega)$:

$$G_1(\omega)=e^{-j\omega T/2}(e^{j\omega T/2}-e^{-j\omega T/2})=2je^{-j\omega T/2}\sin(\omega T/2)\tag{6}$$

Using (6) you can rewrite $G(\omega)$ as

$$G(\omega)=2e^{-j\omega T/2}\frac{\sin(\omega T/2)}{\omega}=
Te^{-j\omega T/2}\frac{\sin(\omega T/2)}{\omega T/2}\tag{7}$$ 

Squaring (7), using $\omega=2\pi f$ and $\text{sinc}(x)=\sin(\pi x)/(\pi x)$ will give you the final form of the result.