We can check for the max case to get the upper bound.
X^{3} will be max when all the remaining 28 digits are 9.
So, 47(99....9_{28 times}) = 4*10^{29} + 7*10^{28}+ 9*10^{27} + 9*10^{26}+.....9*10^{0} =40*10^{28} +7*10^{28}+9(10^{27}+....+10^{0}) =
$47*10^{28} +9 * \frac{10^{28}-1}{10-1} =47*10^{28}+ 9 * \frac{10^{28}-1}{9} =47*10^{28}+10^{28}-1 = 48*10^{28}-1$
Now X^{3} = (48*10^{28})^{3} -3*(48*10^{28})^{2} + 3*48*10^{28}-1 [(a-b)^{3}= a^{3} -3a^{2}b+3ab^{2}-b^{3}]
The a^{3} is the most dominating one.
48^{3}=103823 and 10^{28*3}=10^{84}
So, 103823*10^{84} has 6+84=90 digits.