Since aluminum has a density of #2.7 g//(cm)^3#, #1275g# of aluminum has a volume of #472.2 cm^3#. This is calculated by the formula linking density to mass and volume, #density = mass//volume#.

Given that #volume = area xx depth#, where volume of the aluminum foil is #472.2cm^3# and the area of the foil is (#18.5xx10^3#)#cm^2#, it follows that the depth, or the thickness, of the foil should be equal to #volume//area#.

The volume and the area should all be expressed in the same units, which is why I converted the #m^2# to #cm^2#. As you require the thickness in #mm#, the volume and area should be converted to #mm^3# and #mm^2# respectively.

So you have an aluminum foil of volume (#472.2xx10^3#)#mm^3# and area (#18.5xx10^4#)#mm^2#. You need the thickness of such a foil, so you substitute the values in the formula #depth = volume//area#, which follows as #depth = (472.2xx10^3)//(18.5xx10^4)# #=> 2.55 mm# #~~ 2.6mm#