Let $f$ be a analytic function on $\bar B(0,R)$,with $\| f(z)\|\leq M$ in $\| z \| \leq R$,suppose $\|f(0)\|=a>0$,the the number of zeros of $f$ in $B(0,\frac{1}{3}R)$ is equal or less than $\log 2 \log(\frac{M}{a})$.
This result seems too strong to me. As it ensures that the number of zeros can be controlled by the upper modulus bound and positive bound of modulus at $0$. I assume this is the special property of analytic functions on complex space as no similar result in the real case.
However I've no clue how to prove this one, while the hint suggests that consider the following function: $g(z)=f(z)\prod_{k=1}^n(1-\frac{z}{z_k})^{-1}$,where $\{z_{k}\}$ are the zeros of $f$ in $B(0,\frac{1}{3}R)$
This problem is related to Jensen's formula; in fact solving this problem brings you halfway to the proof of Jensen's formula, in a sense.
Following the hint, you should compute an upper bound for $\max_{|z|\le R} |g(z)|$ (what's the largest it can be, given the locations of the $z_k$?). I obtain $$ \max_{|z|\le R} |g(z)| \le M 2^{-N}, $$ where $N$ is the number of zeros of $f$ in $B(0,R/3)$.
The maximum modulus principle tells you that $|g(0)| \le \max_{|z|\le R} |g(z)|$, and so in particular $|g(0)| = a$ must be smaller than the upper bound. The resulting inequality should solve your problem for you.