Maximum value of a $30$ digit number $ = \underbrace{9999\ldots9}_{\text{30 9's}}\approx 10^{30}$
Now, $10^{30} \leq 2^n$
Taking $\log$ on both sides, we get $\displaystyle 30 \leq n \log 2 \implies n \geq \lceil \dfrac{30}{\log 2}\rceil = 100.$
Minimum value of a $30$ digit number $ = \underbrace{100\ldots0}_{\text{29 0's}}= 10^{29}$
Now, $10^{29} \leq 2^n$
Taking $\log$ on both sides, we get $29 \leq n \log 2 \implies n \geq \lceil \dfrac{29}{\log 2}\rceil = 97.$
So, depending on the $30$ digit number, the binary representation will need minimum $97$ digits and maximum $100$ digits.
Best option, C.