To solve this problem, we need to calculate the electrostatic energy stored in the system after the capacitors are connected.$\\$Given:$\\$• Initial capacitance $C_1 = 900 \,$pF$= 900 \times 10^{-12} \,$F$\\$• Voltage of the battery $V = 100 \,$V$\\$Initially, the energy stored in the charged capacitor is given by:$\\ U_1 = \frac{1}{2} C_1 V^2 \\$Substitute the known values:$\\ U_1 = \frac{1}{2} \times 900 \times 10^{-12} \times (100)^2 \\ U_1 = \frac{1}{2} \times 900 \times 10^{-12} \times 10000 \\ U_1 = 4500 \times 10^{-12} \\ U_1 = 4.5 \times 10^{-9} \,$J$\\$When the charged capacitor is connected to an uncharged capacitor of the same capacitance,$\\$the total capacitance becomes $C_{total} = C_1 + C_2 = 900 \,$pF$+ 900 \,$pF$= 1800 \,$pF$. \\$The charge on the initially charged capacitor is $Q = C_1 V. \\ Q = 900 \times 10^{-12} \times 100 \\ Q = 90000 \times 10^{-12} \\ Q = 9 \times 10^{-9} \,$C$\\$The voltage across the combined capacitors is $V' = \frac{Q}{C_{total}}. \\ V' = \frac{9 \times 10^{-9}}{1800 \times 10^{-12}} \\ V' = \frac{9}{1800} \times 10^{3} \\ V' = 5 \,$V$\\$The energy stored in the system after connection is:$\\ U_2 = \frac{1}{2} C_{total} (V')^2 \\ U_2 = \frac{1}{2} \times 1800 \times 10^{-12} \times (5)^2 \\ U_2 = \frac{1}{2} \times 1800 \times 10^{-12} \times 25 \\ U_2 = 22500 \times 10^{-12} \\ U_2 = 2.25 \times 10^{-9} \,$J$\\$Therefore, the electrostatic energy stored by the system is $2.25 \times 10^{-9} \,$J$. \\$This corresponds to Option 3.