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

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

K_b = 1.53 x 10¯⁴