You must write the balanced reaction: 3 CsBr + Al(OH)3 = AlBr3 + 3 Cs(OH)3
Since MM of Al Br3 is 266 you obtain 25g/266 g/mol = 0,094 mol of salt.
from the balanced reaction, you can see that these mol come from 0,094 x 3 = 0,282 mol of CrBr . Being the molarity N°mol/V you have 0,282 mol/0,125 mL = 2,25 M.
However The reaction is impossible because Al(OH)3 is not soluble while AlBr3 is soluble, and also the amounts of reagents are too high