d
If we approximate the angle \(\theta_{2}\) as \(30^o\) initially then answer will be closer to \(57000\). but if we solve thoroughly, answer will be close to \(55000\) .

So both the answers must be awarded. Detailed solution as following.

Exact solution

By Snell's law  \(1.\sin {40^o} = (1.31)\sin {\theta _2}\)

\(\sin \theta_{2}=\frac{64}{1.31}=\frac{64}{131} \approx .49\)

Now \(\tan {\theta _2} = \frac{{64}}{{\sqrt {{{(131)}^2} - {{(64)}^2}} }}\) \( = \frac{{64}}{{\sqrt {13065} }} \approx \frac{{64}}{{114.3}} = \frac{{\text{d}}}{x}\)

Now number of reflections

\(=\frac{2 \times 64}{114.3 \times 20 \times 10^{-6}}=\frac{64 \times 10^{5}}{114.3}\)

\(=55991 \approx 55000\)

Approximate solution

By Snell's law \(1 . \sin 40^{\circ}=(1.31) \sin \theta_{2}\)

\({\sin \theta_{2}=\frac{0.64}{1.31}=\frac{64}{131} \approx 0.49}\)

If assume  \( \Rightarrow {\theta _2} \approx {30^o}\)

\(\tan {30^o} = \frac{{\text{d}}}{{\text{x}}} \Rightarrow {\text{x}} = \sqrt 3 {\text{d}}\)

Now number of reflections

\(=\frac{\ell}{\sqrt{3} \mathrm{d}}=\frac{2}{\sqrt{3} \times 20 \times 10^{-6}}=\frac{10^{5}}{\sqrt{3}}\)

\(\approx 57735 \approx 57000\)