Explanation

This is a volume-conservation (melting and recasting) problem. The lead cube's volume is 44³ = 85184 cm³. Each bullet has diameter 4 cm (radius 2 cm), so its volume is (4/3)π(2³) = 32π/3 cm³. Using π = 22/7, one bullet occupies 704/21 cm³, so the number of bullets is 85184 ÷ (704/21) = 2541. The subtlety is using the diameter correctly to get radius 2. Hence option (a).