60. Since $a, b, c, d$ are positive real numbers, $a^{2}+b^{2}+c^{2}+d^{2}=1 \geqslant 4 \sqrt[4]{a^{2} b^{2} c^{2} d^{2}}$, so $a b c d \leqslant$ $\frac{1}{16}$. Also,
$$\begin{array}{l} a b+b c+c d+a d+a c+b d \leqslant \frac{a^{2}+b^{2}}{2}+\frac{b^{2}+c^{2}}{2}+\frac{c^{2}+d^{2}}{2}+\frac{a^{2}+a^{2}}{2}+ \\ \frac{a^{2}+c^{2}}{2}+\frac{b^{2}+d^{2}}{2}=\frac{3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}{2}=\frac{3}{2} \end{array}$$

Therefore,
$$\begin{array}{l} a^{2} b^{2} c d+a b^{2} c^{2} d+a b c^{2} d^{2}+a^{2} b c d^{2}+a^{2} b c^{2} d+a b^{2} c d^{2}= \\ a b c d(a b+b c+c d+a d+a c+b d) \leqslant \frac{1}{16} \times \frac{3}{2}=\frac{3}{32} \end{array}$$