Correct Answer: 3 : 1

Solution :
There are two towers A and B
The angles of elevation of the tops of A and B are 30° and 60°, respectively
$\angle$ AOC = 30°, $\angle$ BOD = 60° 
OC = OD
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$
Let AC = h
Then, BD = H
In triangle AOC,
$\tan 30^\circ = \frac{\text{h}}{\text{OC}}$
$\frac{1}{\sqrt{3}} = \frac{\text{h}}{\text{OC}}$ -----(1)
In triangle BOD,
$\tan 60^\circ = \frac{\text{H}}{\text{OD}}$
 $\frac{\sqrt{3}}{1} = \frac{\text{H}}{\text{OD}}$ -----(2)
By Dividing eq (1) and eq (2)
$\frac{1}{3} = \frac{\text{h}}{\text{H}}$
⇒ H : h = 3 : 1 
Then the ratio of the height of tower B to tower A = 3 : 1
Hence, the correct answer is 3 : 1.