Answer: B

Solution:

Write $$\begin{aligned} I_k &= 10^{k+2} + 64 \\ &= 2^{k+2} \cdot 5^{k+2} + 2^6. \end{aligned}$$

If $k \lt 4,$ the first term has fewer than $6$ factors of $2,$ so $N(k) = k + 2 \lt 6.$

If $k \gt 4,$ the first term has at least $7$ factors of $2$ while the second has exactly $6,$ so their sum has exactly $6:$ $N(k) = 6.$

If $k = 4,$ then $I_4 = 2^6(5^6 + 1).$ Since $5^6 + 1$ $= (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big)$ $= 26 \cdot 601,$ it contributes exactly one more factor of $2.$ Thus $N(4) = 7.$

The maximum value is $N(4) = 7.$

Thus, the correct answer is B.