Answer :
The question has already provided a formula to determine the width (W). The question also provides the thickness as well as the number of times (n) the paper would be folded. Hence, we have;
[tex]\begin{gathered} W=\pi\times T\times2\frac{3(n-1)}{2} \\ W=3.14\times0.004\times2\times\frac{3(13-1)}{2} \\ W=0.02512\times\frac{3(12)}{2} \\ W=0.02512\times18 \\ W=0.45216\text{ inches} \end{gathered}[/tex]The notebook paper would have to be 0.45216 inches wide
The formula has also been provided that would help in calculating the length of a long rectangular piece of paper. We've been given the thickness of the paper and the number of times it (n) it would be folded. Therefore, we now have;
[tex]\begin{gathered} L=\frac{\pi T}{6}(2^n+4)(2^n-1) \\ L=\frac{3.14\times0.002}{6}(2^{12}+4)(2^{12}-1) \\ L=0.00104667(4096+4)(4096-1) \\ L=0.00104667(4100)(4095) \\ L=17573.066\text{ inches} \end{gathered}[/tex]