Correct : A

The number of digits in a number. If a number N has d digits, then:
10^d-1 ≤ N < 10^d

For finding digits in X³, I can use logarithms. The number of digits in X³ is:
Number of digits = floor(log₁₀(X³)) + 1
Number of digits = floor(3 × log₁₀(X)) + 1

Since X is a 30-digit number starting with 47, I can estimate its range:
• Minimum value: X ≥ 47 × 10²⁸ (smallest 30-digit number starting with 47)
• Maximum value: X < 48 × 10²⁸ (before it becomes 48...)

The minimum case:
X = 47 × 10²⁸
log₁₀(X) = log₁₀(47) + 28
log₁₀(47) ≈ 1.672 (since log₁₀(47) is between log₁₀(10) = 1 and log₁₀(100) = 2)

So, log₁₀(X) ≈ 1.672 + 28 = 29.672

Therefore:
Number of digits in X³ = floor(3 × 29.672) + 1
= floor(89.016) + 1
= 89 + 1
= 90

Verifying with the maximum case:
X < 48 × 10²⁸
log₁₀(48) ≈ 1.681
log₁₀(X) < 1.681 + 28 = 29.681
Number of digits = floor(3 × 29.681) + 1 = floor(89.043) + 1 = 89 + 1 = 90

Both cases give me 90 digits.