Let `R` is the radius of 1-rupee coin and `r` is the radius of 10 paise coin.
We can create a diagram with the given details.
If we join centers of all 1-rupees coin, it will form an equilateral triangle `ABC`.
Please refer to video for the diagram.
If `O` is the center of 10 paise coin, then,
`/_OAB = /_OBA = 60/2 = 30^@`
Here, `AB = 2R and OA = r+R`.
So, we can say that,
`2(r+R)cos30^@ = 2R`
`(r+R)/R = 2/sqrt3`
`r/R + 1= 2/sqrt3=> r/R = 2/sqrt3-1`
`r/R = (2-sqrt3)/sqrt3 = (2sqrt3-3)/sqrt3`, which is the required ratio.