pH of a diprotic acid?

The acid-dissociation constants of sulfurous acid (H2SO3) are Ka1 = 1.7*10^(-2) and Ka2 = 6.4*10^(-8) at 25.0 °C. Calculate the pH of a 0.163 M aqueous solution of sulfurous acid.

I tried following the example in my book, but my answer didn't match any of the choices.

Ka1 = [H+][HSO3-]/[H2SO3] = x²/(0.163 - x) = 1.7*10^(-2)

→ x ≈ 0.0448221 M

Ka2 = [H+][SO3^2-]/[HSO3-] = (0.0448221 + y)(y)/(0.163 - 0.0448221) = 6.4*10^(-8)

→ y ≈ 1.68742*10^(-7) M

[H+] = x + y = 0.0448221 + 1.68742*10^(-7) = 0.0448222687 M

pH = -log[H+] = -log(0.0448222687) = 1.35

However, my choices are:

A) 1.30

B) 1.86

C) 4.53

Could someone please clarify?