number of moles of KMNO4 = M* V
                                                         Â
= 0.371 M * 0.0245 L
                                                         Â
= 9.0895*10^-3 mol
From reaction,
2 mol of KMNO4 requires 10 mol of HI
so, number of moles of HI required = (10/2)* number of moles of
KMNO4
                                                                       Â
= 5 * 9.0895*10^-3 mol
                                                                       Â
= 0.04545
number of moles of HI required = M*V
0.04545 = 0.204 * V
V=0.223 L
  = 223 mL
Answer: 223 mL