3. The diameter of the base of a cylinder is $4 R$, and its height is $42 R$. The maximum number of spheres with a radius of $R$ that it can hold is ( ).
(A) 28
(B) 56
(C) 58
(D) 60
3. C.

The arrangement is with two on the left and right sides, and two higher ones in the front, placed in layers that cross each other. The distance between successive layers equals the distance between two opposite edges of a regular tetrahedron with the four ball centers as vertices, which is \( d = \sqrt{2} R \). Therefore, \(\frac{42 K - 2 R}{\sqrt{2} R} = 20 \sqrt{2} \approx 28.28\). Hence, the required number is \([28.28 \times 2] + 2 = 58\).