(b)

Given,

$\begin{aligned} M^2=& 2^{66}-2^{46}+2^{32}+2^{30}-2^{16}+1 \\ M^2=& 2^{46}\left[\frac{2^{14}-1}{2-1}\right]+2^{32}+2^{16}\left[\frac{2^{14}-1}{2-1}\right]+1 \\ M^2=& 2^{46}\left[1+2+2^2+\ldots+2^{13}\right] \\ & \quad+2^{32}+2^{16}\left[1+2+2^2+\ldots+2^{13}\right]+1 \\ M^2=& \frac{2^{59}+2^{58}+\ldots+2^{46}}{14 \text { terms }}+2^{32} \end{aligned}$

$\underbrace{+2^{29}+2^{28}+\ldots+2^{16}}_{14 \text { terms }}+2^0$

Therefore, on base $2$ representation of $M^2$, there will be $30$ times digit $1$ .