To find mL of Al needed, start by finding grams of Al that are being used in this reaction:
(3.85mol Al)(27.0g/mol Al)= 104g Al needed.
Since density is m/v, and we know the density and the mass, we can find the volume:
2.70g/mL = (104.g)/mL
Isolate mL:
(104.g)/(2.70g/mL) = 38.5mL needed which is answer C :)