Here n =10,
`r =0.20 mm = 2xx10^(-4)m,`
`R = 14.6cm = 14.6xx10^(-2)m.`
Let d be the diameter of oil molecule
= thickness of layer
volume of olive oil = area xx thickness of layer
`n xx(4)/(3) pi r^3 = piR^2 xxd`
`d = (4)/(3)n (r^3)/(R^2) = (4)/(3) xx(10(2xx10^(-4))^(3))/((14.6xx10^(-2))^2)`
`=5xx10^(-9)m`