Excess pressure `=(2sigma)/(R)=(2xx465xx10^(-1))/(3xx10^(-3))=310Pa`
Total pressure `=1.01xx10^(5)+(2sigma)/(R)`
`=1.01xx10^(5)+310=1.0131xx10^(5)Pa`
Since data is correct up to three significant figures, we should write total pressure inside the drop as `1.01xx10^(5)Pa`.