To understand this question, let's take a simple example.
Three digits smallest decimal number is 100 and the greatest number 999. To represent 100 in binary we need 7 bits, and 10 bits to represent largest number 999.
To represent 1024($2^{10})$, we need $log_{2}^{2^{10}}$ = 10 bits.
To represent 100 in binary, we need $log_{2}^{10^{2}}$ = 2*$log_{2}^{10}$
To represent smallest possible decimal number with d digits in binary : (d-1)*$log_{2}^{10}$ bits
To represent largest possible decimal number with d digits in binary : (d)*$log_{2}^{10}$ bits
Smallest decimal number with 25 digits will require 24*$log_{2}^{10}$ = 80 bits
Largest decimal number with 25 digits will require 25*$log_{2}^{10}$ = 84 bits
PS: consider ciel value.