Correct. Given $\displaystyle{4}\pi{{r}}^{{2}}={100}\usepackage{cmbright}\Huge\$ , you know $\displaystyle{{r}}^{{2}}=\frac{{100}}{{{4}\pi}}\usepackage{cmbright}\Huge\$ , so $\displaystyle\sqrt{{{{r}}^{{2}}}}=\sqrt{{\frac{{100}}{{{4}\pi}}}}=\sqrt{{\frac{{25}}{\pi}}}\usepackage{cmbright}\Huge\$ . Therefore $\displaystyle{r}=\sqrt{{\frac{{25}}{\pi}}}\usepackage{cmbright}\Huge\$ , or $\displaystyle\frac{{5}}{{\sqrt{{\pi}}}}\usepackage{cmbright}\Huge\$ .