Given, $7x^2-13x+a=0$

For rational roots $D\ge 0$ and $D$ must be perfect square.

$\Rightarrow 169-28a\ge 0$

Now, we see from options that $a=7 \& a=8$ are not possible as then $D<0$.

For $a=5$, $D=168-28\times 5=29\ge 0$ But it is not perfect square.

For $a=6$, $D=169-28\times 6=1\ge 0$ and it is a perfect square.

Hence, the smallest possible value of $a$ is $6$.