Step 1: Work with the ratio of internal energies instead of computing $\Delta T$ first.
The rms speed of a gas satisfies $v_{rms}^2 \propto T$. If the rms speed doubles, $v_{rms}^2$ becomes 4 times as large, so the temperature also becomes 4 times as large: $T_2 = 4T_1$. Since the internal energy of a monatomic ideal gas is $U = \frac{3}{2}nRT$, and $U \propto T$ too, this means $U_2 = 4U_1$.
Step 2: Express the heat added as a multiple of the initial internal energy.
\[ Q = U_2 - U_1 = 4U_1 - U_1 = 3U_1 \]
This avoids computing $\Delta T = T_2-T_1$ as a separate step, since the answer is just three times the starting internal energy.
Step 3: Compute the initial internal energy $U_1$.
Taking the initial temperature as $T_1 = 300$ K and $n=2$ mol:
\[ U_1 = \frac{3}{2}nRT_1 = \frac{3}{2}\times2\times8.314\times300 = 3\times8.314\times300 = 7482.6 \ \text{J} \]
Step 4: Multiply by 3 to get the heat added.
\[ Q = 3\times7482.6 = 22447.8 \ \text{J} \approx 22.4 \ \text{kJ} \]
Final Answer:
\[ \boxed{22.4 \ \text{kJ}} \]