Step 1: The key symmetry trick.
Binomial coefficients are symmetric: $^{n}C_r = {}^{n}C_{n-r}$. So a sum looks the same if we replace $r$ by $n-r$. We will use this to link $b_n$ with $a_n$.
Step 2: Write the same sum backwards.
In $b_n = \sum \frac{r}{{}^{n}C_r}$, replace $r$ with $n-r$. The bottom stays the same by symmetry, only the top changes: \[ b_n = \sum_{r=0}^{n} \frac{n-r}{{}^{n}C_r} \]
Step 3: Add the two forms of $b_n$.
Add the original and the reversed version term by term: \[ 2b_n = \sum_{r=0}^{n} \frac{r + (n-r)}{{}^{n}C_r} \]
Step 4: Simplify the top.
The top becomes just $n$, a constant: \[ 2b_n = n \sum_{r=0}^{n} \frac{1}{{}^{n}C_r} \]
Step 5: Recognise $a_n$.
That leftover sum is exactly $a_n$. So \[ 2b_n = n\, a_n \]
Step 6: Form the ratio.
Divide both sides by $2a_n$: \[ \boxed{ \frac{b_n}{a_n} = \frac{n}{2} } \]