Start with the balanced chemical equation for the equilibrium reaction that exists when hydroxylamine reacts with water
#NH_2OH_((aq)) + H_2O_((l)) rightleftharpoons NH_3OH_((aq))^(+) + OH_((aq))^(-)#
Use the solution's pH to determine the pOHm which will allow you to calculate the concentration of the hydroxide ions
#pH_"sol" = 14 - pOH => pOH = 14 - pH_"sol"#
#pOH = 14 - 10.11 = 3.89#
The concentration of hydroxide ions will be
#[OH^(-)] = 10^(-pOH) = 10^3.89 = 1.29 * 10^(-4)"M"#
Since you have #1:1# mole ratios between all the species of interest, the initial concentration of hydroxylamine will decrease by the same amount the concentration of hydroxide increased.
At equilibrium, you'll have
#[NH_3OH^(+)] = [OH^(-)] = 1.29 * 10^(-4)"M"#
#[NH_2OH] = [NH_2OH]_0 - [OH^(-)] = 0.15 - 1.29 * 10^(-4) = "0.14987 M"#
By definition, the base dissociation constant, #K_b#, will be equal to
#K_b = ([OH^(-)] * [NH_3OH^(+)])/([NH_2OH])#
#K_b = (1.29 * 10^(-4) * 1.29 * 10^(-4))/(0.14987) = color(green)(1.1 * 10^(-9))#
