Let the mass be placed at a distance x from the left end.
Let the tension in the left string be T1 and tension in the right string be T2.
Now taking moments about the left end,
48x+12*0.2=T2*0.4....(1) (Since weight of the rod acts at the midpoint of the rod.)
Taking moments about right end,
48(0.4-x)+12*0.2=T1*0.4......(2)
Also given f=1/2Lroot(T1/m)=2/2Lroot(T2/m)
So T1=4T2.....(3)
Solve (1),(2) and (3) you will get x=5cm
So the mass should be kept at a distance of 5cm from the left end.