Here, radius of oil drop `= r = (0.6xx10^(-3))/(2) m = 0.3xx10^(-3)m`
`R =12cm =12xx10^(-2)m`
Thickness of oil film ` = ("Volume of oil drop")/("Area of film") = ((4)/(3)pir^3)/(piR^3)`
t=(4/3pi(0.3xx10^(-3))^(3))/(piR^(2)) = (4(0.3xx10^(3))^(3))/(3xx(12x10^(-2))^2) = (4xx0.027xx10^(-5))/(3xx12xx12) m = 2.5xx10^(-9)m`.
If we assume that the film is one molecule thick, then molecular size of oleic acid `~= 2.5xx10^(-9)m`