How to calculate Ksp using solubility

Can some one please help me work this out. Many thanks.
The solubility of lead(II) fluoride in water is 0.0175%(w/v) at 29 degrees C. Calculate the Ksp of this salt at this temperature.
a) 1.46 x 10^-12 M
b) 1.46 X10^-9M
c) 3.64 X10^-10M
d)1.31 x 10^-9M
e) 1.31 x 10^-12M
f) 3.28 x 10^-10M

0.0175%(w/v) = 0.01775 g/100 mL = 0.175 g PbF2 / L

0.175 g PbF2 / L * (mol PbF2)/(245.2 g PbF2) = 0.0007137 mol/L

PbF2 --> Pb + 2F

When it dissolves:

[Pb] = 0.0007137 mol/L
[F] = 0.0014274 mol/L

Ksp = [Pb][F]^2

Ksp = (0.0007137)(0.0014274)^2 ==> Ksp = 1.45 * 10^-9

The answer is B