Step 1: What controls a lens focal length.
For one fixed lens the surfaces stay the same, so the only thing that changes with colour is the refractive index. The lens maker formula tells us $\frac{1}{f} = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, and since the bracket is fixed, $f$ behaves like $\frac{1}{\mu-1}$.
Step 2: Turn this into a simple ratio.
Because $f$ is proportional to $\frac{1}{\mu-1}$, two colours of the same lens are linked by $\frac{f_r}{f_v} = \frac{\mu_v-1}{\mu_r-1}$.
Step 3: Note the given numbers.
Violet: $\mu_v=1.55$, $f_v=20$ cm. Red: $\mu_r=1.50$, $f_r=?$
Step 4: Find each bracket value.
$\mu_v-1 = 0.55$ and $\mu_r-1 = 0.50$.
Step 5: Plug into the ratio.
\[ f_r = f_v\cdot\frac{\mu_v-1}{\mu_r-1} = 20\times\frac{0.55}{0.50} \]
Step 6: Do the arithmetic.
$\frac{0.55}{0.50}=1.1$, so $f_r = 20\times 1.1 = 22$ cm. Red light bends less, so its focal length is longer than that of violet, which matches our answer.
\[ \boxed{f_r = 22\ \text{cm}} \]