On the first road sign the passenger saw the number $10x + y$, $0 < x \le 9$, $0 \le y \le 9$. One hour later he saw the number $100y + x$, $y \ne 0$. During that time the bus has driven $(100y + x) - (10x + y) = 9(11y - x)\text{ km}$. The bus had constant speed so in the next two hours it has driven $2 \cdot 9(11y - x) = 18(11y - x)$ and the passenger saw the number $100y + 10z + x$, $0 < z \le 9$. So we get $18(11y - x) = (100y + 10z + x) - (100y + x)$ or $9(11y - x) = 5z$. The last equality is possible only if $z = 9$ and $11y - x = 5$. Because $0 < x, y \le 9$ we obtain $x = 6$ and $y = 1$. So we get the speed from $9(11y - x) = 1 \cdot v$, or $v = 45\text{ km/h}$. The numbers that the passenger saw on the road signs are $61$, $106$ and $196$.