smc

If $D(n)$ is even, then the binary expansion of $n$ will both begin and end with a $1$, because all positive binary numbers begin with a $1$, and if you switch digits twice, you will have a $1$ at the end. Thus, we are only concerned with the $49$ odd numbers between $1$ and $98$ inclusive. All of these odd numbers will have an even $D(n)$. $D(n)=0$ will be given by the numbers $1,11,111,1111,11111,111111$, which is a total of $6$ numbers. We skip $D(n)=2$ for now, and move to $D(n)=4$, which is easier to count. The smallest $D(n)=4$ happens when $n=10101$. To get another number such that $D(n)=4$, we may extend any of the five blocks of zeros or ones by one digit. This will form $110101,100101,101101,101001,101011$, all of which are odd numbers that have $D(n)=4$. To find seven digit numbers that have $D(n)=4$, we can again extend any block by one, so long as it remains less than $1100001$ or under. There are five cases. 1) Extending $110101$ is impossible without going over $1100001$. 2) Extending $100101$ by putting a $1$ at the beginning will go over $1100001$, but the other four extensions work, giving $1000101,1001101,1001001,1001011$. 3) Extending $101101$ by putting a $1$ at the beginning will go over $1100001$, but the other four extensions give $1001101,1011101,1011001,1011011$. However, $1001101$ already appeared in #2, giving only three new numbers. 4) Extending $101001$ at the first group is impossible. The other four extensions are $1001001,1011001,1010001,1010011$, but the first two are repeats. Thus, there are only two new numbers. 5) Extending $101011$ at the first group is impossible. The other four extensions give $1001011,1011011,1010011,1010111$, but only the last number is new. Thus, there is $1$ five digit number, $5$ six digit numbers, and $4+3+2+1=10$ seven digit numbers under $1100001$ for which $D(n)=4$. That gives a total of $16$ numbers. There smallest number for which $D(n)=6$ is $1010101$, which is under $98$. Further extensions, as well as cases where $D(n)>6$, are not possible. Thus, we know that there are $6$ odd numbers that have $D(n)=0$, and $16$ odd numbers that have $D(n)=4$, and $1$ number that has $D(n)=6$. The remaining odd numbers must have $D(n)=2$. This means there are $49-6-16-1=26$ numbers that have $D(n)=2$, which is option $\boxed{26}$