If each edge of a cube is increased by 10%, then the percentage increase in its surface area is:

Let the edge of the cube = $$a$$

Surface area of the cube = $$6a^2$$

Edge of the cube if increased by 10% = $$\frac{110}{100}\times a=\frac{11}{10}a$$

Surface area of the cube after increase = $$6\left(\frac{11}{10}a\right)^2=6\left(\frac{121}{100}a^2\right)$$

$$\therefore\ $$Percentage increase Surface = $$\frac{6\left(\frac{121}{100}\right)a^2-6a^2}{6a^2}\times100$$

$$=\frac{\frac{21}{100}a^2}{a^2}\times100$$

$$=\frac{21}{100}\times100$$

$$=21\%$$

Hence, the correct answer is Option D