Answer: 100 suns
Step-by-step explanation:
We can solve this with the following relation:

Where:
is the diameter of a dime
is the diameter of the Sun
is the distance between the Sun and the pinhole
is the amount of dimes that fit in a distance between the sunball and the pinhole
Finding
:


This is roughly the diameter of the Sun
Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

So, we have to divide this distance between
in order to find how many suns could it fit in this distance:
