Here's a simple approach purely based on observation:
Note that $47^{3}$ is $103823$ (6 digits)
Start with a smaller number, say $4700$ : It's cube will be $103823$ followed by $000000$.
How about $4799$? It's cube will be $110522$ followed by $894400$. Did number of digits increase? No.
Trying with more number of digits after 47, we can see that the first part $47^{3}$ will always give 6 digits and the remaining number of digits after it are getting tripled in number, eg for two digits after $47$ we got six digits in the cube of $4700$ as shown above.
So extending it for $47$ followed by $28$ more digits, we will get $6$ digits for $47^{3}$ and $3*28$ more digits for the number after $47$.
We get total : $6 + 84 = 90$ digits.
Option A.