With the Fourier transform we obtain
$$ w(\mathbf{x},t)=\int_{-\infty}^{\infty}d\mathbf{k} \exp{\left[i\mathbf{kx}\right]}\frac{1}{2\pi}\exp{\left[-(D\mathbf{k}^2t)\right]}= \frac{1}{\sqrt{4\pi Dt}}\exp{\left[-(\mathbf{x}^2/4Dt)\right]}, $$with the normalization condition
$$ \int_{-\infty}^{\infty}w(\mathbf{x},t)d\mathbf{x}=1. $$