The value of $\left\{\frac{3^{2003}}{28}\right\}=\frac{\mathrm{a}}{\mathrm{b}},\{.\}$ denotes the fractional part, then (b-a) is equal to.

$3^{2003}=3^{2001}\cdot 3^2=9(27{)}^{667}=9(28-1{)}^{667}$

$=9\left({ }^{667}{\mathrm{C}}_0{28}^{667}{-}^{667}{\mathrm{C}}_1(28{)}^{666}+\dots .{-}^{667}{\mathrm{C}}_{667}\right)$

that means if we divides 3²⁰⁰³ by 28, remainder is 19.

thus, $\left\{\frac{3^{2003}}{28}\right\}=\frac{19}{28}=\frac{\mathrm{a}}{\mathrm{b}}\Rightarrow \mathrm{b}-\mathrm{a}=28-19=9$