Correct Answer - `99%`
Redox changes are :
`FetoFe^(+2)+2r^(-) " " ("in" H_2SO_4)`
`Fe^(+2)to Fe^(+3)+e^(-) " " ("with" K_2Cr_2O_7)`
`6e+Cr_2^(+8)to2Cr^(+3)`
m.eq. of `Fe^(+2)` in 20 mL =m.eq. of `K_2Cr_2O_7=30xx1/30=1`
`:.` m.eq of `Fe^(+2)` in 100 mL `=(1xx100)/20=5`
`:.` m.moles of `Fe^(+2)=(meq)/(v.f.)=5/1=5`=m.moles of Fe
`:.` Mass of pure Fe in wire =`0.28/0.2828xx100=99%`