Explanation :
Fixed end moment
$ {\mathrm M}_\mathrm{FQP}={\mathrm M}_\mathrm{FQR}=0 $
Writing slope deflection equation of QP and QR
$ \begin{array}{l}{\mathrm M}_\mathrm{Qp}=0+\frac{2\mathrm{EI}}5(2{\mathrm\theta}_\mathrm Q)=\frac45{\mathrm{EIθ}}_\mathrm Q\\{\mathrm M}_\mathrm{QR}=\frac{2\mathrm{EI}}4(2{\mathrm\theta}_\mathrm Q)={\mathrm{EIθ}}_\mathrm Q\\\end{array} $
As $ {\mathrm M}_\mathrm{Qp}+{\mathrm M}_\mathrm{QR}=-180\;\mathrm{kN}.\mathrm m $
$ \begin{array}{l}\Rightarrow\frac45{\mathrm{EIθ}}_\mathrm Q+{\mathrm{EIθ}}_\mathrm Q=-180\\\Rightarrow{\mathrm\theta}_\mathrm Q=\frac{-180\times5}{9\times10^4}=-0.01\;\mathrm{rad}\end{array} $
Hence, $ {\mathrm\theta}_\mathrm Q=0.01\;\mathrm{rad} $
Alternative Solution :
$ \begin{array}{l}\frac{4\mathrm{EIθ}}5={\mathrm M}_1\\\frac{4\mathrm{EIθ}}4={\mathrm M}_2\\{\mathrm M}_1+{\mathrm M}_2=180\\\Rightarrow1.8\mathrm{EIθ}=180\\\mathrm\theta=\frac{100}{\mathrm{EI}}=\frac{100}{10^4}=0.01\;\mathrm{rad}.\end{array} $