A

Step 1: Use Torricelli's theorem to find the velocity of the water exiting the hole without extra pressure. The velocity is given by \( v = \sqrt{2g h} \), where \( h = 10 \text{ m} \) and \( g = 10 \text{ m/s}^2 \).
Step 2: Calculate velocity:
\( v = \sqrt{2 \cdot 10 \cdot 10} = \sqrt{200} = 10\sqrt{2} \text{ m/s} \).
Step 3: The horizontal range \( R \) when water exits the hole is given by \( R = \frac{v t}{g} \), where \( t \) is the time of flight which can be simplified to \( t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{20}{10}} = 2 \text{ s} \).
Thus, \( R = 10\sqrt{2} \cdot 2 = 20\sqrt{2} \).
Step 4: To achieve a range of \( 2R \), the velocity must double, thus \( v' = 2v = 20\sqrt{2} \text{ m/s} \).
Step 5: The new velocity under extra pressure is given by \( v' = \sqrt{2(g+h')}, \text{ where } h' = h + \frac{P_{extra}}{\rho g} \). Therefore, \( 20\sqrt{2} = \sqrt{2\left(10 + \frac{P_{extra}}{1000 \cdot 10}\right)} \).
Step 6: Squaring both sides gives: \( 800 = 10 + \frac{P_{extra}}{1000} \).
Step 7: Rearranging provides: \( 790 = \frac{P_{extra}}{1000} \) or \( P_{extra} = 790000 ext{ Pa} \).
Step 8: Convert this to atm: \( \frac{790000}{100000} = 7.9 ext{ atm} \).
Therefore, the extra pressure required is approximately \( 7.9 ext{ atm} \). Thus, the correct answer is A.