Let decimal representation of $a$ consist of exactly $k$ digits, and binary representation consists of exactly $l$ times more digits, i.e. of $kl$ digits. Since all digits of $a$ equal to either $0$ or $1$, then the number $a$ should belong to, from one side, the interval $[10^{k-1}, \frac{1}{6}(10^k - 1)]$, and from the other side, to the interval $[2^{k/l-1}, 2^{k/l} - 1]$. Hence we come to two inequalities: $10^{k-1} \le 2^{k/l} - 1$ and $\frac{1}{6}(10^k - 1) \ge 2^{k/l-1}$. After transformation, we get $\frac{2}{3} < (\frac{10}{27})^k < 10$. This inequality, obviously, is wrong for $l \ge 4$.

For $l=1$ we have the only possible value $k=1$ and the first answer - number $1$.

If $l=2$, then $k=2$. From all two-digit numbers, having only $0,1$ as digits, only $10$ suits - the second answer.

For the last case, $l=3$, possible values are $k \in \{7,8,9,10\}$. Let's examine them.

For $k=10$ we have $10^9 > 2^{29} + 2^{28}$. That is, if $a$ is a ten-digit number such that written thrice it represents binary form of itself, then it has to have $1$ in the second-highest decimal position, i.e. $a > 11 \cdot 10^8 > 2^{30}$ - a contradiction (binary representation of $a$ should be a $30$-digit number).

For all other values of $k$ we see that $a = (2^{2k} + 2^k + 1)$. For $k=9$, $2^{2k} + 2^k + 1 = 262657$. Numbers $10^0, 10^1, ..., 10^8$ give remainders $1, 10, ..., 100000, 212029, 19034$ and $190340$ under the division by $262657$, respectively. Since $a$ is a sum of several powers of $10$ with the highest $10^8$, we have to choose from $1, 10, ..., 100000, 212029, 19034, 190340$ such that they sum up either to $262657-190340$, or to $2 \cdot 262657-190340$. By examination of options we see that it is impossible to get such numbers. Similarly, we can check all other cases. Let's see (for example) the case $k=8$. $2^{10} + 2^8 + 1 = 65793$. The remainders of $10^0, 10^1, ..., 10^7$ under the division by $65793$ equal $1, 10, ..., 10000, 234207, 13105$ and $-536$. We have to select a combination of powers, summing up to $536$ - others impossible. By checking two last digits ($00, 01, 10, 07, 05$) we see that it is impossible to get the required two last digits $36$.