If I am not wrong, the equation
b == x - ss*InverseCDF[NormalDistribution[0, 1], b][[1]]
should be equivalent to
2 b == Erfc[(b - x)/(Sqrt[2] ss)]
and this one Mathematica can do:
b[x_, ss_] :=
Solve[2 b == Erfc[(b - x)/(Sqrt[2] ss)] && 0 <= b <= 1, b, Reals]
b[3/10, 1/10]