0.040 mol / dm³. (2 sig. fig.)
[tex](\text{CH}_3)_3\text{N}[/tex] in this question acts as a weak base. As seen in the equation in the question, [tex](\text{CH}_3)_3\text{N}[/tex] produces [tex]\text{OH}^{-}[/tex] rather than [tex]\text{H}^{+}[/tex] when it dissolves in water. The concentration of [tex]\text{OH}^{-}[/tex] will likely be more useful than that of [tex]\text{H}^{+}[/tex] for the calculations here.
Finding the value of [tex][\text{OH}^{-}][/tex] from pH:
Assume that [tex]\text{pK}_w = 14[/tex],
[tex]\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}[/tex].
[tex][\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}[/tex].
Solve for [tex][(\text{CH}_3)_3\text{N}]_\text{initial}[/tex]:
[tex]\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}[/tex]
Note that water isn't part of this expression.
The value of Kb is quite small. The change in [tex](\text{CH}_3)_3\text{N}[/tex] is nearly negligible once it dissolves. In other words,
[tex][(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}[/tex].
Also, for each mole of [tex]\text{OH}^{-}[/tex] produced, one mole of [tex](\text{CH}_3)_3\text{NH}^{+}[/tex] was also produced. The solution started with a small amount of either species. As a result,
[tex][(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}[/tex].
[tex]\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}[/tex],
[tex][(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}[/tex],
[tex][(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}[/tex].
