Simple na

$\frac{1}{3}+\frac{1}{50}<1$ and $1<\frac{1}{3}+\frac{50}{50}<2$ , so we just have to find where $[\frac{1}{3}+\frac{n}{50}]=1$

To get $[\frac{1}{3}+\frac{n}{50}]=1$ , we must have $[\frac{1}{3}+\frac{n}{50}]>1$ . From there we have $n>33$

The maximum value of $n$ is $50$ , so there are $17$ values of $n$ such that $[\frac{1}{3}+\frac{n}{50}]=1$ . All other values of $n$ give $[\frac{1}{3}+\frac{n}{50}]=0$ , so $E=17$

$17!$ has $8$ multiple of $2$ , $4$ multiples of $4$ , $2$ multiples of $8$ , and $1$ multiple of $16$ .

Thus, the exponent of two is $8+4+2+1=\boxed{15}$