 Solution

For an isotropic three-dimensional ideal walk, each Cartesian component contributes $N{b}^2/3$ to the mean square. Keeping two components gives $⟨R_{\mathrm{proj}}^2⟩=2N{b}^2/3$. The planar projection of an isotropic random vector therefore has $R_{\mathrm{proj},\mathrm{rms}}=b\sqrt{2N/3}$, smaller than the unprojected planar ideal walk by $\sqrt{2/3}$. The corresponding two-dimensional Gaussian mean distance has the same ratio.