The amount of toilet paper left will be proportional to the area of the paper on the cross-section of the roll. The cross sections of the two rolls are shown below.

On the new roll, this area is $\pi\times6^2-\pi\times2^2=36\pi-4\pi=32\pi$ cm$^2.$

On my roll, this area is $\pi\times3^2-\pi\times2^2=9\pi-4\pi = 5\pi$ cm$^2.$

So $\frac{5}{32}=15.625\%$ of the toilet paper is left.

On the new roll, this area is $\pi\times6^2-\pi\times2^2=36\pi-4\pi=32\pi$ cm$^2.$

On my roll, this area is $\pi\times3^2-\pi\times2^2=9\pi-4\pi = 5\pi$ cm$^2.$

So $\frac{5}{32}=15.625\%$ of the toilet paper is left.

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