Initially, just before release, the system has potential energy $U$:

$$T=U=mgL$$

As the string hits the horizontal bar, it enters into a different 'orbit', this one with radius $L-H$. At the highest point of this loop, the mass is at height $2(L-H)$. Assuming full conservation of energy, then the new total energy is:

$$T=2mg(H-L)+K$$

Where $K$ is the kinetic energy at the top of the loop. So:

$$mgL=2mg(H-L)+K$$
So that:

$$K=mg(3L-2H)$$

With $v$ the tangential velocity at the top of the loop:

$$K=\frac12 mv^2=mg(3L-2H)$$

So:

$$v^2=2g(3L-2H)$$

The centripetal force $F_c$ is:

$$F_c=\frac{mv^2}{R}=\frac{mv^2}{L-H}$$

For there not to tension in the string at the top of the loop:

$$mg=\frac{mv^2}{L-H}$$

$$g=\frac{2g(3L-2H)}{L-H}$$

So I get:

$$L-H=2(3L-2H)$$

$$H=\frac53 L$$