= Solution
Use the standard <orientations> on the three <spheres> and the complex <orientation> on <Complex projective space>. Let $h\in H^2(\mathbb{CP}^3;\mathbb Z)$ satisfy $\langle h^3,[\mathbb{CP}^3]\rangle=1$. The <cohomology ring of complex projective space> is $\mathbb Z[h]/(h^4)$. The <Künneth theorem> gives
$$
H^*((S^2)^3;\mathbb Z)=\mathbb Z[a,b,c]/(a^2,b^2,c^2),
\qquad |a|=|b|=|c|=2,
\qquad \langle abc,[(S^2)^3]\rangle=1.
$$
Write the <induced map on cohomology> as $f^*h=pa+qb+rc$ with integers $p,q,r$. Naturality and <graded commutativity of the cup product> give
$$
f^*(h^3)=(pa+qb+rc)^3=6pqr\,abc.
$$
The <mapping degree> is therefore $d=6pqr$. This is the <factorial degree obstruction for products of two-spheres>. Every positive value is a positive multiple of six, so \b[the least possible positive degree is at least $6$].
To attain it, identify each oriented <sphere> with $\mathbb{CP}^1$. Send three projective linear forms to the coefficients of their product:
$$
F:(\mathbb{CP}^1)^3\longrightarrow\mathbb{CP}^3,
\qquad ([a_1:b_1],[a_2:b_2],[a_3:b_3])\longmapsto[c_0:c_1:c_2:c_3],
$$
where
$$
\prod_{i=1}^3(a_i z+b_i w)=c_0z^3+c_1z^2w+c_2zw^2+c_3w^3.
$$
No product of nonzero linear forms is the zero polynomial. Rescaling any factor rescales all coefficients by the same nonzero scalar. The map is therefore well-defined and is a <holomorphic map> on the products of projective charts.
Choose a homogeneous cubic with three distinct roots, all in an affine chart. Its preimages are exactly the $3!=6$ orderings of its three projective linear factors. In root coordinates, the derivative of the elementary-symmetric coefficient map has <determinant> equal, up to sign, to the <Vandermonde determinant>, the product of the three pairwise root differences. This is nonzero for distinct roots, so all six preimages are regular. A nonsingular complex-linear derivative preserves the underlying real <orientation>, since its real <determinant> is the squared absolute value of its complex <determinant>. Each preimage consequently contributes $+1$ to the <mapping degree>. Hence
$$
\boxed{\deg F=6,\qquad d_{\min}=6.}
$$
This <polynomial multiplication map to complex projective space> provides the requested example without assuming that a prescribed <cohomology> homomorphism is realizable by a continuous map.
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