Answer
${\bf 2.0\times 10^8}\;\rm m/s$
Work Step by Step
We have here a double-slit experiment.
We know that the spacing between fringes in a double-slit interference experiment is given by
$$\Delta y=\dfrac{\lambda L}{d}\tag 1$$
where $L$ is the distance between the double-slit and the screen, $d$ is the slit separation, and $\lambda$ is the wavelength of the neutrons which we can find by using de Broglie wavelength's formula.
$$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}$$
Solving for $v$ since the author asks for neutrons speed;
$$v=\dfrac{h}{m\lambda}\tag 2$$
Solving (1) for $\lambda$;
$$\lambda=\dfrac{\Delta y \;d}{ L}$$
Plug it into (2);
$$v=\dfrac{hL}{m \Delta y \;d} $$
Plug the known;
$$v=\dfrac{(6.63\times 10^{-34})(3.5)}{ \Delta y(0.1\times 10^{-9})(1.67\times 10^{-27})} \tag 3$$
As we see, the only unknown here is $\Delta y$ which we can find by using the 7 peaks on the given graph and the 100-$\mu$m scale.
It seems that the distance between the first peak and the 7th peak is $6\Delta y$ which is about 4 times the given scale.
Noting that this is a rough estimation.
Thus,
$$6\Delta y=400\;\rm \mu m$$
$$\Delta y=\bf 66.7\;\rm \mu m$$
Plug into (3);
$$v=\dfrac{(6.63\times 10^{-34})(3.5)}{ (66.7\times 10^{-6})(0.1\times 10^{-9})(1.67\times 10^{-27})} $$
$$v=\color{red}{\bf 2.0\times 10^8}\;\rm m/s$$
