Correct Answer - B
`pH =3 , log [(1)/(H^(+))] =3`
`[(1)/(H^(+)]] = " Antilog " 3=10^(3)`
`[H^(+)] =10^(-3) , " Molarity "=0.1M`
`H^(+) sqrt(K_(a) C) " or " (10^(-3)) = sqrt(K_(a)C)`
`10^(-6) =K_(a) xx c " or " K_(a) = (10^(-6))/(c)`
`K_(a) =(10^(-6))/(0.1) =10^(-5)`