1. Use the relation between speed and refractive index

For any medium, $n = \frac{c}{v} \quad \Rightarrow \quad v = \frac{c}{n}$

So, $v_A = \frac{c}{n_A}, \qquad v_B = \frac{c}{n_B}$

Given: $v_A - v_B = 2.6 \times 10^7\ \text{m/s}$ $n_B = 1.47, \qquad c = 3 \times 10^8\ \text{m/s}$

2. Find the speed in medium B

$v_B = \frac{3 \times 10^8}{1.47}$

$v_B \approx 2.0408 \times 10^8\ \text{m/s}$

3. Find the speed in medium A

Since $v_A - v_B = 2.6 \times 10^7$ we get $v_A = v_B + 2.6 \times 10^7$

$v_A \approx 2.0408 \times 10^8 + 0.26 \times 10^8 = 2.3008 \times 10^8\ \text{m/s}$

4. Find refractive index of medium A

$n_A = \frac{c}{v_A} = \frac{3 \times 10^8}{2.3008 \times 10^8}$

$n_A \approx 1.304$

5. Find the required ratio

We need: $\frac{n_B}{n_A} = \frac{1.47}{1.304}$

$\frac{n_B}{n_A} \approx 1.127 \approx 1.13$

6. Match with options

The correct option is: $\boxed{\text{C: }1.13}$