A slightly different path to the [OH¯]

pH = 11.68, so pOH = 14 - pH = 14.00 - 11.68 = 2.32

[OH¯] = 10¯pOH = 10¯2.32 = 4.7863 x 10¯3 M

[HB+] = [OH¯] = 4.7863 x 10¯3 M

[B] = 0.15

Kb = [(4.7863 x 10¯3) (4.7863 x 10¯3)] / 0.15

Kb = 1.53 x 10¯4