= Solution
For the shorter uniform rod, the <parallel axis theorem> gives $I=m(a^2/12+x^2)$. Its equivalent <simple pendulum> length is $I/(mx)=x+a^2/(12x)$. The rod of length $2a$ pivoted at an end has equivalent length $2(2a)/3=4a/3$. Matching the <periods> therefore requires
$$
x+\frac{a^2}{12x}=\frac{4a}{3},\qquad 12x^2-16ax+a^2=0,
$$
$$
x=\frac{a}{6}(4\pm\sqrt{13}).
$$
A pivot on the shorter rod requires $0<x\le a/2$, leaving
$$
\boxed{x=\frac{4-\sqrt{13}}6a\simeq0.0657415a.}
$$
Either side of the rod's <centre of mass> gives this distance. The other algebraic root is beyond the rod and would require a support extension of negligible <mass>.
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