Let the radius of the rod be \(r\) cm The length of the rod is \(h = 14\) cm So, the volume of the rod = \(\pi r^2 \times 14\) cubic cm Volume of each sphere = \(\frac{4}{3} \pi \times 8^3\) cubic cm According to the condition: \[ \pi r^2 \times 14 = 21 \times \frac{4}{3} \pi \times 8^3 \] Simplifying: \[ r^2 \times 14 = 7 \times 4 \times 8 \times 8 \times 8 \Rightarrow r^2 = \frac{7 \times 4 \times 8 \times 8 \times 8}{14} \Rightarrow r^2 = 2 \times 8 \times 8 \times 8 \Rightarrow r = \sqrt{2 \times 2 \times 2 \times 2 \times 8 \times 8} \Rightarrow r = 32 \] Therefore, the radius of the cross-section of the rod was 32 cm.