Find volume of ammonia gas required to prepare a solution of given pH
The $\mathrm{pH}$ of your final ammonia solution is $11.0$. When you dissolve ammonia in water it tends to ionize with water to give following equilibrium:
$$\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-} \quad K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = 1.8 \times 10^{-5}$$
Suppose final concentration of $\ce{NH3}$ is $c$, and equlibrium concentrations of $\ce{NH4+}$ and $\ce{NH4+}$ are $\alpha$. Thus:
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{\alpha \cdot \alpha}{c - \alpha} = 1.8 \times 10^{-5}$$
$$\alpha^2 + 1.8 \times 10^{-5}\alpha - 1.8 \times 10^{-5}c =0 \tag1$$
Since, $\mathrm{pOH}$ of the solution is $3.0$ ($14 - \mathrm{pH}$), $[\ce{OH-}] = \alpha = 1.00 \times 10^{-3}$, from equation $(1)$:
$$(1.00 \times 10^{-3})^2 + 1.8 \times 10^{-5}\times 1.00 \times 10^{-3} = 1.8 \times 10^{-5}c $$
$$1.8 \times 10^{-5}c = 1.00 \times 10^{-6} + 1.8 \times 10^{-8} = 1.018 \times 10^{-6} $$
$$\therefore \ \ c = \frac{1.018 \times 10^{-6}}{1.8 \times 10^{-5}} = \pu{5.66 \times 10^{-2} molL-1} $$
This is exactly what you got without showing your work. However, you made only $\pu{100 mL}$ of solution. Thus, amount of ammonia dissolved in $\pu{100 mL}$ of water is:
$$$\pu{0.100 L} \times \pu{5.66 \times 10^{-2} molL-1} = \pu{5.66 \times 10^{-3} mol}$
Assuming gaseous ammonia is behaving like ideal gas, you can use $PV = nRT$ here to find the volume of $\pu{5.66 \times 10^{-3} mol}$ of ammonia at STP:
$$V= \frac{nRT}{P} = \frac{(\pu{5.66 \times 10^{-3} mol})(\pu{0.082 LatmK-1mol-1})(\pu{298 K})}{\pu{1 atm}} = \pu{0.138 L}$$