1. First phase: free fall for $50\,\text{m}$

The parachutist starts from rest and falls freely without air resistance for $50\,\text{m}$.

Using $v^2=u^2+2gs$ with $u=0,\quad g\approx 10\,\text{m/s}^2,\quad s=50\,\text{m}$ we get $v^2=0+2(10)(50)=1000$ so the speed just before opening the parachute is $v=\sqrt{1000}=10\sqrt{10}\,\text{m/s}.$

2. Second phase: motion with parachute open

After the parachute opens, the parachutist is decelerated at $a=-2\,\text{m/s}^2$ (taking downward as positive, retardation means acceleration upward, hence negative).

He reaches the ground with speed $v=3\,\text{m/s}.$

Let the distance covered after the parachute opens be $s$.

Again use $v^2=u^2+2as$ where $u^2=1000,\quad v^2=9,\quad a=-2.$

Thus, $9=1000+2(-2)s$ $9=1000-4s$ $4s=991$ $s=247.75\,\text{m}.$

3. Total height from which he bailed out

Total height $H=50+247.75=297.75\,\text{m}.$

This is closest to $293\,\text{m}.$

Hence the correct option is D.

4. Note on approximation

If one uses $g=9.8\,\text{m/s}^2$, the result is $u^2=2(9.8)(50)=980,$ $9=980-4s,$ $s=242.75\,\text{m},$ $H=50+242.75=292.75\,\text{m}\approx 293\,\text{m}.$

So with the standard precise value $g=9.8\,\text{m/s}^2$, the answer is exactly option D.