The rest distance between Lucknow and Bhopal is 1000 km . A nonstop...

The rest distance is $L_0 = 1000 \text{ km} = 10^6 \text{ m}$.

The speed of the train is $v = 360 \text{ km h}^{-1} = 360 \times \dfrac{5}{18} \text{ m s}^{-1} = 100 \text{ m s}^{-1}$.

According to the length contraction formula, the distance in the train's frame is $L = L_0 \sqrt{1 - \dfrac{v^2}{c^2}}$.

Since $v \ll c$, using the binomial expansion gives L \approx L_0 \left(1 -\dfrac{v^2}{2c^2}\right).

The decrease in length is $\Delta L = L_0 - L =$\dfrac{L_0 v^2}{2c^2}.

Substituting the values:

$\Delta L =$\dfrac{10^6 \times (100)^2}{2 \times (3 \times 10^8)^2}

$\Delta L =$\dfrac{10^{10}}{18 \times 10^{16}}

$\Delta L =$\dfrac{1}{18} \times$10^{-6} \text{ m}$

$\Delta L \approx 0.0555 \times 10^{-6} \text{ m} = 55.5 \text{ nm} \approx 56 \text{ nm}$

Thus, the distance in the train's frame is $56 \text{ nm}$ less than $1000 \text{ km}$.