### Solution:

It is given that,The size of the paper is 30 cm by 18 cm.

(Cylinder 1) By rolling lengthwise.

Height of cylinder 1 (h1) = 18 $\n \n cm\n $ , diameter of the cylinder 1 = 30 cm

Since the circumference of the base of the cylinder is $\n \n 2\pi r\n $ . So, $\n \n \n \n 2\pi \n r\n 1\n \n =30\n \n \n \n \n \u21d2\n r\n 1\n \n =\n \n 30\n \n 2\pi \n \n \n \n \n \n \n \u21d2\n r\n 1\n \n =\n \n 30\xd77\n \n 2\xd722\n \n \n \n \n \n \n \u21d2\n r\n 1\n \n =\n \n 105\n \n 22\n \n cm\n \n \n \n \n $ Thus, the volume (V1) is,

$\n \n \n \n =\pi \n r\n 1\n \n \n \n 2\n \n \n h\n 1\n \n \n \n \n \n \n =\n \n 22\n 7\n \n \xd7\n \n 105\n \n 22\n \n \xd7\n \n 105\n \n 22\n \n \xd718\u2004c\n m\n 3\n \n \n \n \n \n \n $ (Cylinder II) By rolling breadthwise. Height of cylinder 2 $\n \n (\n h\n 2\n \n )=30cm\n $ , diameter of cylinder 2 = 18 $\n \n cm\n $

Since the circumference of the base of the cylinder is $\n \n 2\pi r\n $ . So,

$\n \n \n \n 2\pi \n r\n 2\n \n =18\n \n \n \n \n \u21d2\n r\n 2\n \n =\n \n 18\n \n 2\pi \n \n \n \n \n \n \n \u21d2\n r\n 2\n \n =\n \n 63\n \n 22\n \n cm\n \n \n \n \n $ Thus, the volume (V2) is,

$\n \n \n \n =\pi \n r\n 2\n \n \n \n 2\n \n \n h\n 2\n \n \n \n \n \n \n =\n \n 22\n 7\n \n \xd7\n \n 63\n \n 22\n \n \xd7\n \n 63\n \n 22\n \n \xd730\u2004c\n m\n 3\n \n \n \n \n \n \n $ Now, divide $\n \n \n V\n 1\n \n \n $ by $\n \n \n V\n 2\n \n \n $ .

$\frac{{V}_{1}}{{V}_{2}}=\frac{\frac{22}{7}\times \frac{105}{22}\times \frac{105}{22}\times 18}{\frac{22}{7}\times \frac{63}{22}\times \frac{63}{22}\times 30}$

$=\frac{105\times 105\times 18}{63\times 63\times 30}$

$=\frac{198450}{119070}$

$=\frac{5}{3}$ Thus, the ratio of the two cylinders is $\n \n 5:3\n $ .

Therefore, option 4 is correct.