The correct option is C 0.111.......
Given operation is (0.333....)2
Now, 0.333....=3×0.111...............(1)
Let 'x' be 0.111....
Multilpy x with 10.
⇒ 10x = 1.11....
= 1 + 0.111....
= 1 + x (∵x=0.111....)
⇒ 9x = 1
⇒ x = 19
∴ 0.111.... = 19................(2)
Substituting the value of 0.111... in (1) we get,
0.333....=3×19
=13
∴(0.333....)2=(13)2
=19
∵19=0.111.... [from eq (2)]
⇒(0.333....)2=0.111.....