Correct Answer: 72.8%

Solution : Given: The radius and the height of the cone are each increased by 20%.
Let the radius and height of the cone be 10 m.
When radius and height is increased by 20%, then new height and radius = $\frac{120}{100}\times10$ = 12 m
Volume of cone before increment $=\frac{1}{3}\pi r^2 h$
$=\frac{1}{3}\pi \times 10^2 \times 10 =\frac{1000}{3}\pi$
Volume of cone after increment $=\frac{1}{3}\pi r^2 h$
$=\frac{1}{3}\pi \times 12^2 \times 12 =\frac{1728}{3}\pi$
Change in percentage = $\frac{\text{Volume after increment}-\text{Volume before increment}}{\text{Volume before increment}}\times100$
= $\frac{\frac{1728}{3}\pi-\frac{1000}{3}\pi}{\frac{1000}{3}\pi}\times 100$
= $\frac{\frac{(1728-1000)}{3}\pi}{\frac{1000}{3}\pi}\times 100$
= $\frac{728}{1000}\times 100$
= $72.8$%
Hence, the correct answer is 72.8%.