Let the side of the rhombus = a
Area of the triangle containing the diagonal = $\frac{1}{2}$ times area of the rhombus

One of the diagonal = 6 cm

Semi perimeter = $\frac{6+a+a}{2}=a+3$

Now the area of the triangle containing the diagonal =$\sqrt{s(s-a)(s-b)(s-c)}$ 

$3\sqrt{7}$ = $\sqrt{(a+3)(a+3-a)(a+3-a)(a+3-6)}$
                     =  $\sqrt{(a+3)(a-3)\left(9\right)}$
                     = $\sqrt{(a^2-9)\left(9\right)}$
We take square on both sides and get,
63 = $(a^2-9)\times 9$
$\Rightarrow (a^2-9)$ = 7
$\Rightarrow a^2$ = 7+9 = 16

$\Rightarrow$ a= 4cm