The density of aluminum is 2.70x10^3 km/m^3, the density of iron is 7.86x10^3 kg/m^3. Find the radius of a solid iron sphere that balances a solid aluminum sphere of radius 2.86cm on an equal-arm balance. my work: (4/3)(pi)(.286m)^3 = .000098m^3 for volume X weight = .264576kg .264576/7.86x10^3 = .000034 (4/3)(pi)(RADIUS)^3 = .000034 Radius = .020029945725539 blah blah METERS-- can someone help check this for me? (remember to covert cm to m!)

EDIT: nvm, not as hard as i thought when i first glanced at it...but still dont know the answer cause im too lazy to figure it out...