Solution:

The number $N$ satisfies $$7\cdot10^{312}\lt N\lt 8\cdot10^{312}.$$ Multiplying or dividing by a power of $10$ only shifts the decimal point, so we only need the leading factor left after taking out the largest convenient power of $10$.

For $r=2$, $\sqrt N$ has leading factor between $\sqrt7$ and $\sqrt8$, so $f(2)=2$.

For $r=3$, the leading factor is between $\sqrt[3]{7}$ and $\sqrt[3]{8}$, so $f(3)=1$. For $r=4$, the leading factor is between $\sqrt[4]{7}$ and $\sqrt[4]{8}$, so $f(4)=1$.

For $r=5$, since $312=5\cdot62+2$, the leading factor is between $\sqrt[5]{700}$ and $\sqrt[5]{800}$. Since $3^5\lt700\lt800\lt4^5$, $f(5)=3$.

For $r=6$, the leading factor is between $\sqrt[6]{7}$ and $\sqrt[6]{8}$, so $f(6)=1$. The sum is $2+1+1+3+1=8$.

Thus, the answer is A .