\(n_{CaO}=\dfrac{3,6.10^{23}}{6.10^{23}}=0,6\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0,04\left(mol\right)\)
\(CaO\left(0,06\right)+H_2O--->Ca\left(OH\right)_2\left(0,06\right)\)
\(\Rightarrow n_{Ca\left(OH\right)_2}\left(sau\right)=0,06+0,04=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)