Answer
$r=1.10\times10^{-10}m$.
Work Step by Step
See Figure 29-15. Each nitrogen atom is a distance r/2 from the center of mass.
$$I=2m_N\left(\frac{1}{2}r \right)^2=\frac{1}{2}m_Nr^2$$
Express the characteristic rotational energy of $N_2$ about its center of mass.
$$E_{rot}=\frac{\hbar ^2}{2I}$$
Combine the previous 2 expressions, and then find the bond length.
$$E_{rot}=\frac{\hbar ^2}{2(\frac{1}{2})m_Nr^2}$$
$$r=\frac{\hbar}{\sqrt{E_{rot}m_N}}$$
$$=\frac{1.055\times10^{-34}J \cdot s}{\sqrt{(2.48\times 10^{-4}eV)(1.60\times10^{-19}J/eV)(14.01u)(1.66\times10^{-27}kg/u)}}$$
$$=1.10\times10^{-10}m$$