A rectangular paper sheet of dimensions $54\text{ cm} \times 4\text{ cm}$ is taken. The two longer edges of the sheet are joined together to create a cylindrical tube. A cube whose surface area is equal to the area of the sheet is also taken. Then, the ratio of the volume of the cylindrical tube to the volume of the cube is

1. Find the volume of the cylindrical tube:

   • The sheet dimensions are $54\text{ cm} \times 4\text{ cm}$.

   • When the two longer edges of length $54\text{ cm}$ are joined together, the height of the resulting cylinder is $h = 54\text{ cm}$, and the circumference of the circular cross-section is the shorter edge, $C = 4\text{ cm}$.

   • The radius $r$ of the cylinder is given by:

     $$2\pi r = 4 \implies r = \frac{2}{\pi}\text{ cm}$$

   • The volume of the cylindrical tube is:

     $$V_{\text{cylinder}} = \pi r^2 h = \pi \left(\frac{2}{\pi}\right)^2 (54) = \pi \times \frac{4}{\pi^2} \times 54 = \frac{216}{\pi}\text{ cm}^3$$

2. Find the volume of the cube:

   • Area of the rectangular sheet $= 54 \times 4 = 216\text{ cm}^2$.

   • The total surface area of a cube with side length $a$ is $6a^2$.

   • Given $6a^2 = 216\text{ cm}^2$, we have:

     $$a^2 = 36 \implies a = 6\text{ cm}$$

   • The volume of the cube is:

     $$V_{\text{cube}} = a^3 = 6^3 = 216\text{ cm}^3$$

3. Calculate the ratio:

   $$\frac{V_{\text{cylinder}}}{V_{\text{cube}}} = \frac{\frac{216}{\pi}}{216} = \frac{1}{\pi}$$

Hence, the correct option is (A).