The total surface area of a cube is $\(6 \times(\text { Side })^2\)$, so the total surface area of cube having side/edge 1 units each would be $\(6 \times(1)^2=6\)$

The length of rectangle inscribed in the cube is the hypotenuse length of the right triangle in which one perpendicular side is the edge of the cube and the other perpendicular side is half the edge of the cube i.e. the perpendicular sides are 1 unit \& $\(\frac{1}{2}\)$ unit. So, using Pythagoras theorem i.e. Hypotenuse $\({ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$, we get the length of the rectangle as

$\(\sqrt{(1)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}\)$

Next the width of the rectangle inscribed in the cube is same as the side of the cube i.e. 1 unit.

Thus, the area of the rectangle is Length $\(\times\)$ Width $\(=\frac{\sqrt{5}}{2}\times 1=\frac{\sqrt{5}}{2}\)$


Finally the ratio of the total surface area of the cube to the area of the rectangle is $\(\frac{6}{\frac{\sqrt{5}}{2}}=\frac{12}{\sqrt{5}}\)$

Hence the answer is (B).