Correct. Given $\displaystyle{T}=\pi\sqrt{{\frac{{L}}{{245}}}}\usepackage{cmbright}\Huge\$ , we know that $\displaystyle\frac{{T}}{\pi}=\sqrt{{\frac{{L}}{{245}}}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{{\left(\frac{{T}}{\pi}\right)}}^{{2}}={{\left(\sqrt{{\frac{{L}}{{245}}}}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ . This means $\displaystyle\frac{{{T}}^{{2}}}{{\pi}^{{2}}}=\frac{{L}}{{245}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{L}=\frac{{{245}{{T}}^{{2}}}}{{\pi}^{{2}}}\usepackage{cmbright}\Huge\$ .