**Answer**

(4)

**Explanation**

Let $\alpha$ and $4 \beta$ be roots of $x^{2}-6 x+a=0$ and

$\alpha, 3 \beta$ be the roots of $x^{2}-c x+6=0$, then $\alpha+4 \beta=6$ and $4 \alpha \beta=a$

$\alpha+3 \beta=c$ and $3 \alpha \beta=6$

we get $\alpha \beta=2 \Rightarrow a=8$

So the first equation is $x^{2}-6 x+8=0 \Rightarrow x=2,4$ If $\alpha=2$ and $4 \beta=4$ then $3 \beta=3$

If $\alpha=4$ and $4 \beta=2$, then $3 \beta=3 / 2$ (non-integer)

$\therefore$ common root is $x=2$