1. Convert speeds into SI units

Train A: $108\ \text{km/h} = 108\times \frac{5}{18} = 30\ \text{m/s}$

Train B: $72\ \text{km/h} = 72\times \frac{5}{18} = 20\ \text{m/s}$

2. Time taken by each train to completely cross the tunnel

To completely cross a tunnel, a train must cover: $\text{distance} = \text{length of tunnel} + \text{length of train}$

So,

For train A (length $l$): $t_A = \frac{L+l}{30}$

For train B (length $4l$): $t_B = \frac{L+4l}{20}$

Given that train A takes $35\,$s less time than train B: $t_B - t_A = 35$

Thus, $\frac{L+4l}{20} - \frac{L+l}{30} = 35$

3. Use the given relation $L=60l$

Substitute $L=60l$: $\frac{60l+4l}{20} - \frac{60l+l}{30} = 35$

$\frac{64l}{20} - \frac{61l}{30} = 35$

Take LCM $60$: $\frac{192l - 122l}{60} = 35$

$\frac{70l}{60} = 35$

$\frac{7l}{6} = 35$

$l = 35\times \frac{6}{7} = 30\ \text{m}$

4. Find tunnel length

$L = 60l = 60\times 30 = 1800\ \text{m}$

5. Check options

$L=1800\ \text{m}$

So the correct option is C.