Number of significant figures when going from base 10 to binary.

For fixed $N_{10}$, no such $N_2$ exists.

Proof:
Given a tentative $N_2$, set $n = 2^{N_2} + 1$ (two $1$s with $(N_2 - 1)$ $0$s between). Then in base $2$, $n\cdot 2^k$ always rounds up to $(n+1)\cdot 2^k$. However $n\cdot 2^k - 1$ rounds down to $(n-1)\cdot 2^k$. Since $n\cdot 2^k$ and $n\cdot 2^k - 1$ round to the same $N_{10}$ significant digits in base $10$ for at least one $k$, $N_2$ does not work.