We need to know the area of the cross section of the sample before anything else. In this case, the aluminum sample has a width of 30.00 centimeters and a thickness of 15.0 micrometers,
Area = .30 x .000015 = 4.5 x 10^-6 m^2
the tensile strength to be 311 MPa so,
311x10^6 = F / 4.5 x 10^-6 m^2
F = 1399.5 N is needed
Hope this answers the question.