When the coin is just on the verge of slipping, it just overcomes the force of static friction (\\\\({ F }_{ s }\\\\)) . which acts as centripetal force for being that coin in the circular motion.So we have \\\\({ F }_{ s }=m{ \\omega }^{ 2 }r\\\\) If we double the \\\\(\\omega\\\\), then \\\\(r\\\\) will also self adjust to a new \\\\({ r }^{ ' }\\\\), since the static friction will remain same. We have \\\\(\\quad m{ \\omega }^{ 2 }r=m{ (2\\omega ) }^{ 2 }{ r }^{ ' }\\\\ \\Rightarrow { r }^{ ' }=\\dfrac { r }{ 4 } \\ cm=\\dfrac { 4 }{ 4 }\\ cm=1\\ cm\\\\)