Dear Sun, how far are you?

Here the width of folded paper is increasing exponentially.

So

$d = w \times 2^{ t }$

where

d = distance to the sun = 1.496x10^11 m

w = width of paper = 0.1 mm = 1x10^-4 m

t = number of paper folds

solving the above equation we get

$\frac { d }{ w } \quad ={ \quad 2 }^{ t }$ $\log _{ 2 }{ \frac { d }{ w } } \quad ={ \quad \log _{ 2 }{ { 2 }^{ t } } }$ $t\quad =\quad \log _{ 2 }{ \frac { d }{ w } }$ $t\quad =\quad \log _{ 2 }{ \frac { 1.496\times { 10 }^{ 11 } }{ 1\times { 10 }^{ -4 } } }$ $t\quad =\quad 50.41\quad \\$

So we can reach to the SUN in approximately 50 folds

Quick and dirty back of the envelope calculation:

Assume the sun is ~100 million miles away from the Earth and that a mile is ~1000 meters (really it's closer to 1600, but since the distance to the sun could be off by more than a factor of 1.6, this is probably fine). Letting $n$ denote the number of folds, our equation is

$2^n \times 10^{-4} \approx 10^8 \times 10^3$

$2^n \approx 10^{15}$

Using the approximation $2^{10}\approx 10^3$ , we get $n \approx 50$ .