Explanation : X = 4777....(7 29times)
It can be written as X = 4.7777... * 1029
X3 = (4.777⋯∗1029)3
= (4.777…)3 ∗ 1087
Now, even if we round up 4.777... to 5, we could represent 53 = 125 in 3 digits. So, can say (4.77...)3also has 3 digits before decimal point.
So, X3 requires 3 + 87 = 90 digits.