 Solution

At fixed flux and bandwidth, brightness is inversely proportional to ${\Sigma }_x{\Sigma }_x^{\prime }{\Sigma }_y{\Sigma }_y^{\prime }$. Lowering both geometric emittances by ten makes the electron rms widths and divergences smaller by $\sqrt{10}$. Add the unchanged radiation widths in quadrature again and divide the old phase-space product by the new one. The resulting brightness factor is approximately $\boxed{3.87}$, an increase of about $287\%$, rather than the factor 100 that would follow if diffraction were negligible in both planes. The exact value under the adopted width convention is given by the accompanying check and plot.