Estimating a sinc period and width of a rectangular pulse
Let's look at what is generally known:

Time domain is generally inversely related to frequency domain in as much as if you have a narrow width in the time domain the frequency domain spectrum will be wider.  and vice versa.  When you have narrower time domain pulses you need higher frequencies to represent it in the frequency domain therefore more bandwidth -> wider plot.

Looking at your sinc function to find when the function will go to zero you simply need to look at the sin aspect of it.  At what values does sin = \$0\$ ?  Answer \$ 0, \pi , 2\pi , 3\pi \$ ...  That means the value \$ {\tau}f = 0, \pi.2\pi,3\pi ... \$

There is one exception here, as \$ {\tau}f\$ -> \$0\$ sinc ->\$1\$ becasue the sin and the \$\tau\$ cancel in the limit. so you only need to look at \$\pi,2\pi,3\pi, ...\$

Lets pick \$\tau=1\$ then the rect function will run from \$ -\frac{\tau}{2}\$ to \$ \frac{\tau}{2}\$ and the first zero in the frequency domain will be at f=1.

Lets pick \$\tau=2\$ then the rect function will run from \$ -\frac{\tau}{1}\$ to \$ \frac{\tau}{1}\$ and the first zero in the frequency domain will be at f=\$\frac{1}{2}\$.

a wider time pulse means a narrower frequency band.

Lets pick \$\tau=\frac{1}{2}\$ then the rect function will run from \$ -\frac{\tau}{4}\$ to \$ \frac{\tau}{4}\$ and the first zero in the frequency domain will be at f=\$\frac{2}{1}= 2\$.

a narrower time pulse means a wider frequency band.