1. Speed of the falling mass just before collision

The $1\,\text{kg}$ body falls freely through height $h=100\,\text{m}$.

Using $v=\sqrt{2gh}=\sqrt{2\cdot 10\cdot 100}=\sqrt{2000}=20\sqrt{5}\,\text{m/s}$

2. Velocity just after collision

The body sticks to the platform of mass $3\,\text{kg}$, so this is a perfectly inelastic collision.

Conserving momentum during collision: $1\cdot (20\sqrt{5})=(1+3)V$ $V=\frac{20\sqrt{5}}{4}=5\sqrt{5}\,\text{m/s}$

So just after collision, the combined mass $4\,\text{kg}$ moves downward with speed $V=5\sqrt{5}\,\text{m/s}$

3. Compression of spring after collision

Now the $4\,\text{kg}$ system compresses the spring further until momentarily coming to rest.

Let the additional maximum compression be $x$.

At the instant just after collision, kinetic energy is $KE=\frac12 (4)(5\sqrt{5})^2=2\cdot 125=250\,\text{J}$

During downward motion by $x$, gravity does positive work: $W_g=Mgx=4\cdot 10\cdot x=40x$

At maximum compression, all this energy is stored in the spring: $\frac12 kx^2=\frac12 (1.25\times 10^6)x^2=6.25\times 10^5 x^2$

Hence, $6.25\times 10^5 x^2 = 250 + 40x$

Since $x$ is small, let us solve: $625000x^2-40x-250=0$

Divide by $5$: $125000x^2-8x-50=0$

Using quadratic formula, $x=\frac{8+\sqrt{64+4\cdot 125000\cdot 50}}{2\cdot 125000}$ $x=\frac{8+\sqrt{25000064}}{250000}$

Since $\sqrt{25000064}\approx 5000$

we get $x\approx \frac{5008}{250000}=0.020\,\text{m}=2\,\text{cm}$

This does not match any option, so we must interpret the spring compression carefully.

4. Correct interpretation: spring already compressed by platform weight

Initially, the $3\,\text{kg}$ platform is resting on the spring, so the spring is already compressed by $x_0=\frac{3g}{k}=\frac{30}{1.25\times 10^6}=2.4\times 10^{-5}\,\text{m}$ which is negligible.

After sticking, the new equilibrium compression for $4\,\text{kg}$ is $x_e=\frac{4g}{k}=\frac{40}{1.25\times 10^6}=3.2\times 10^{-5}\,\text{m}$ also negligible.

So gravity has negligible effect compared to the very stiff spring, and the main compression comes from converting post-collision kinetic energy to spring energy: $\frac12 (4)V^2\approx \frac12 kx^2$ $250=\frac12 (1.25\times 10^6)x^2$ $x^2=\frac{500}{1.25\times 10^6}=4\times 10^{-4}$ $x=2\times 10^{-2}\,\text{m}=2\,\text{cm}$

This still does not match the options.

5. Check if total energy from top is directly used after sticking

Before collision, the falling mass has energy $mgh=1\cdot 10\cdot 100=1000\,\text{J}$

But because the collision is perfectly inelastic, a large part is lost during collision. Only $250\,\text{J}$ remains as kinetic energy afterward. Therefore using $1000\,\text{J}$ would be incorrect.

If one incorrectly used $1000\,\text{J}$: $\frac12 kx^2=1000$ $x=0.04\,\text{m}=4\,\text{cm}$ which gives option B.

6. Conclusion

By proper mechanics including inelastic collision, the compression comes out close to $2\,\text{cm}$, which is not among the options. The stored answer $4\,\text{cm}$ corresponds to ignoring energy loss in the sticking collision.

Therefore, the stored answer appears inconsistent with the stated condition that the body sticks to the platform.