Answer
1.14 mm.
Work Step by Step
The value of $y_{20}$ is much smaller than R. Use the approximate expression $y_m=R\frac{m \lambda}{d}$.
$$d=\frac{20R\lambda}{y_{20}}=\frac{(20)(1.20m)(502\times10^{-9}m)}{10.6\times10^{-3}m}=1.14\times10^{-3}m$$
