[Solution] Let the radius of the circle be $R$, the distance from the center $O$ to the chord $AB$ be $x$, and the distance to the chord $CD$ be $6-x$.
Thus, we have $7^{2}+x^{2}=R^{2}=5+(6-x)^{2}$,
solving this gives $x=1$.
Since $EF$ is equidistant from $AB$ and $CD$, the distance from $O$ to $EF$ is 2.
Therefore, $\left(\frac{\sqrt{a}}{2}\right)^{2}+2^{2}=R^{2}=7^{2}+1^{2}=50$,
solving this gives
$$
a=184 .
$$

Thus, the answer is $(E)$.
$20 \cdot 20$ The lengths of two parallel chords of a circle are 10 and 14, and they are 6 units apart. A chord parallel to and equidistant from these two chords has a length of $\sqrt{a}$, then $a$ equals
(A) 144.
(B) 156.
(C) 168.
(D) 176.
(E) 184.
(46th American High School Mathematics Examination, 1995)