The balanced chemical equation is 2Al+2NaOH+2H2O→2NaAlO2+3H2.
54 g (2 mol) of Al will give 3 moles (or 67.2 L) of hydrogen.
0.15 g of Al will give 67.254×0.15=0.189L of hydrogen.
The volume of hydrogen is at 1 bar and 273 K. The volume at 1 bar and 200C will be, V′=PVT′P′T=1×0.189×2931×273=0.2028L=202.8 ml.