= Solution
Interpret the reported angles as total scattering angles $\psi=2\theta$. The photon <energy> gives
$$
\lambda=\frac{hc}{E_\gamma}=1.23984\times10^{-10}\,\mathrm m.
$$
Combining $2d\sin\theta=\lambda$ with the two index assignments yields
$$
a_{111}=\frac{\sqrt3\lambda}{2\sin11.4^\circ}=0.54323\,\mathrm{nm},\qquad
a_{220}=\frac{\sqrt8\lambda}{2\sin18.8^\circ}=0.54409\,\mathrm{nm}.
$$
The slight difference is consistent with the rounded angular data, giving
$$
\boxed{a\simeq0.544\,\mathrm{nm}.}
$$
The intensity estimate needs a stated convention. The geometric <structure factors> obey $|F_{111}|^2=32|f_{111}|^2$ and $|F_{220}|^2=64|f_{220}|^2$, so equal atomic factors give a \b[single-directed-reflection] ratio $1/2$. For a powder line, the <powder diffraction multiplicities> are eight for $\{111\}$ and twelve for $\{220\}$. Hence a multiplicity-and-structure-factor estimate is
$$
\boxed{\frac{I_{111}}{I_{220}}\simeq\frac{8\cdot32}{12\cdot64}=\frac13}
$$
if other angle-dependent factors are neglected.
A measured integrated powder intensity also contains the <Lorentz-polarization factor>, atomic form factor, thermal-motion factor and instrument/sample corrections. For example, in the common unpolarized symmetric powder geometry,
$$
LP(\theta)\propto\frac{1+\cos^22\theta}{\sin^2\theta\cos\theta},\qquad
\frac{LP(11.4^\circ)}{LP(18.8^\circ)}=2.9174.
$$
Including this factor but still taking equal atomic factors gives $I_{111}/I_{220}\simeq0.9725$, rather than $1/3$. More generally,
$$
\frac{I_{111}}{I_{220}}\simeq0.9725
\left|\frac{f_{111}}{f_{220}}\right|^2e^{-2(W_{111}-W_{220})}
$$
within that geometry, before other corrections. The actual form factors and measurement geometry are not supplied, so no unique precision experimental peak-height ratio follows. The values $1/2$, $1/3$ and $0.9725$ refer to distinct explicitly stated levels of approximation; a powder family must not be silently treated as one directed reflection.
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