A rhombicosidodecahedron is an Archimedean solid, one of thirteen convex isogonal nonprismatic solids constructed of two or more types of regular polygon faces. It has $20$ regular triangular faces, $30$ square faces, and $12$ regular pentagonal faces, as shown below. How many rotational symmetries does a rhombicosidodecahedron have?

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To determine the number of rotational symmetries of a rhombicosidodecahedron, we need to consider the symmetries of the solid that map it onto itself. The rhombicosidodecahedron is a highly symmetric polyhedron, and its symmetries are related to the symmetries of the icosahedron and dodecahedron, which share the same symmetry group.

1. **Identify the symmetry group**:
   The rhombicosidodecahedron has the same rotational symmetry group as the icosahedron and dodecahedron. This group is known as the icosahedral group, denoted by $A_5$, which is the alternating group on 5 elements. The icosahedral group has 60 rotational symmetries.

2. **Calculate the number of rotational symmetries**:
   The icosahedral group $A_5$ has 60 elements. These elements correspond to the rotational symmetries of the rhombicosidodecahedron. Each symmetry can be thought of as a rotation that maps the polyhedron onto itself.

3. **Verification**:
   To verify, consider the faces of the rhombicosidodecahedron:
   - It has 20 triangular faces.
   - It has 30 square faces.
   - It has 12 pentagonal faces.

   Each of these faces can be rotated around their centers, but the overall symmetry of the polyhedron is determined by the icosahedral group. The number of rotational symmetries is not simply the product of the number of faces and their individual rotational symmetries, but rather the total number of distinct rotations that map the entire polyhedron onto itself.

Therefore, the number of rotational symmetries of a rhombicosidodecahedron is given by the order of the icosahedral group, which is 60.

The final answer is $\boxed{60}$.