Hello Adarsh,

According to the law of conservation of energy take the the ground as zero. Take point A as the point at a height H from the ground and point B at a height of h/3.

Now, as the mass m is dropped to h/3 i.e. to point B,  the height from the ground at that point will be H-h/3= 2h/3 .

At this point the potential energy will be mg(2h/3) = 2mgh/3

And the kinetic energy will be = 1/2mv^2

We have to find v at point B

At point A, u= 0 as the mass is just dropped

and a = g , s= h/3

So, by v^2 - u^2 = 2as

            v^2 - 0 = 2×g×h/3

            v^2 = 2gh/3

Now put the value of v^2 in the above equation i.e. 1/2 mv^2 we will get 

Kinetic energy= 1/2m (2gh/3)

                         = mgh/3

Potential energy= 2mgh/3

Now the ratio of potential energy to kinetic energy will be 

          =P.E. / K.E.

          = 2mgh/3 / mgh/3

          = 2 

Hope this will help you. Feel free to ask any query.

All the best!!