Say $a = \overline{d_1d_2\ldots d_k}$. Then $b=a(10^k+1)$ and we need $a^2 \mid b$, i.e. $a\mid 10^k + 1$. But clearly $a<10^k+1<11a$, so $b/a^2 =7$ (since $2,3,5$ do not divide $10^k+1$ for any $k$); the value $7$ is possible when $k\equiv 3\pmod{6}$.