Integration by parts results in $$\frac{\partial \langle x^2\rangle}{\partial t} = -Dxw(x,t)|_{x=\pm\infty}+ 2D\int_{-\infty}^{\infty}w(x,t)dx=2D,$$ leading to $$\langle x^2\rangle = 2Dt,$$ and the variance as $$\begin{equation} \tag{6} \langle x^2\rangle-\langle x\rangle^2 = 2Dt. \end{equation}$$ The root mean square displacement after a time $t$ is then $$\begin{equation*} \sqrt{\langle x^2\rangle-\langle x\rangle^2} = \sqrt{2Dt}. \end{equation*}$$