= Solution
Under a proper <rotation matrix> $R$, the product of coordinates transforms as a rank-$n$ <tensor>. The unit ball is unchanged by the substitution $\mathbf y=R\mathbf x$, whose <Jacobian determinant> is one. Hence
$$
\begin{aligned}
R_{i_1j_1}\cdots R_{i_nj_n}S_{j_1\cdots j_n}
&=\int_B(R\mathbf x)_{i_1}\cdots(R\mathbf x)_{i_n}\,dV_{\mathbf x}\\
&=\int_B y_{i_1}\cdots y_{i_n}\,dV_{\mathbf y}=S_{i_1\cdots i_n}.
\end{aligned}
$$
Thus $S$ is an <isotropic tensor>. An <isotropic second-rank tensor> has the form $S_{ij}=A\delta_{ij}$. Taking the <trace> and using <spherical coordinates>,
$$
3A=\int_B|\mathbf x|^2\,dV=4\pi\int_0^1r^4\,dr=\frac{4\pi}5.
$$
Therefore the <isotropic second moment of a ball> is
$$
\boxed{S_{ij}=\frac{4\pi}{15}\delta_{ij}.}
$$
An <isotropic third-rank tensor> has the form $B\epsilon_{ijk}$, but $S_{ijk}$ is symmetric in every pair of indices, forcing $B=0$. Alternatively, $\mathbf x\mapsto-\mathbf x$ makes its integrand odd on the ball. Thus
$$
\boxed{S_{ijk}=0.}
$$
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