We are given a right circular cone that is sliced by a plane parallel to its base, creating a smaller cone at the top and a frustum below. The volume of the original cone is 27 times that of the smaller cone. Additionally, the lateral surface area of the smaller cone, when unfolded, forms a sector of a circle with a central angle of $120^\circ$. We are tasked with determining the ratio $H/l$, where $H$ is the height of the original cone and $l$ is the slant height of the smaller cone.

Step 1: Set up the relationship for volumes

Let the radius and height of the original cone be $R$ and $H$, respectively. The radius and height of the smaller cone (after slicing) will be $r$ and $h$, respectively. The volume of a cone is given by the formula: $V = \frac{1}{3} \pi r^2 h.$ Thus, the volume of the original cone is: $V_{\text{original}} = \frac{1}{3} \pi R^2 H,$ and the volume of the smaller cone is: $V_{\text{small}} = \frac{1}{3} \pi r^2 h.$ We are told that the volume of the original cone is 27 times that of the smaller cone, so: $V_{\text{original}} = 27 V_{\text{small}}.$ Substituting the expressions for the volumes: $\frac{1}{3} \pi R^2 H = 27 \times \frac{1}{3} \pi r^2 h.$ Canceling the common terms of $\frac{1}{3} \pi$, we get: $R^2 H = 27 r^2 h.$ Since the smaller cone is similar to the original cone (as they are both right circular cones), the corresponding dimensions of the cones are proportional. That is: $\frac{r}{R} = \frac{h}{H} = k \quad \text{for some constant } k.$ Thus, we can express $r = kR$ and $h = kH$. Substituting these into the volume equation: $R^2 H = 27 (kR)^2 (kH),$ which simplifies to: $R^2 H = 27 k^3 R^2 H.$ Canceling $R^2 H$ from both sides (assuming $R, H \neq 0$): $1 = 27 k^3,$ which gives: $k^3 = \frac{1}{27},$ so: $k = \frac{1}{3}.$

Step 2: Relate the lateral surface areas

Next, we use the information about the lateral surface area of the smaller cone. The lateral surface area of a cone is given by: $A = \pi r l,$ where $l$ is the slant height. The slant height of the original cone is $L$, and the slant height of the smaller cone is $l$. We are told that when the lateral surface area of the smaller cone is unfolded, it forms a sector of a circle with a central angle of $120^\circ$.

The lateral surface area of the smaller cone is $A_{\text{small}} = \pi r l$, and the radius of the sector is $l$, so the arc length of the sector corresponds to the circumference of the base of the smaller cone, which is $2\pi r$. The formula for the arc length of a sector is: $\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi l,$ where $\theta = 120^\circ$. Thus: $2\pi r = \frac{120^\circ}{360^\circ} \times 2\pi l,$ which simplifies to: $2\pi r = \frac{1}{3} \times 2\pi l.$ Canceling $2\pi$ from both sides: $r = \frac{1}{3} l.$ Substituting $r = kR = \frac{1}{3} R$, we have: $\frac{1}{3} R = \frac{1}{3} l,$ which gives: $l = R.$

Step 3: Find the ratio $H/l$

We now know that $k = \frac{1}{3}$ and $l = R$. Since $k = \frac{h}{H}$, we have $h = \frac{1}{3} H$. The ratio $H/l$ is: $\frac{H}{l} = \frac{H}{R}.$ Using the fact that $k = \frac{1}{3}$, we find: $\frac{H}{R} = 3.$ Thus, the value of $H/l$ is: $\boxed{3}.$