Given, Rest mass of the proton, $m_0 = 1.67 \times 10^{-27} \text{ kg}$ Speed of light, $c = 3 \times 10^8 \text{ m/s}$ Relativistic mass, $m = 2m_0$ To find the speed $v$ of the proton, we use the relativistic mass formula: $$ \begin{aligned} m &= \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 2m_0 &= \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \sqrt{1 - \frac{v^2}{c^2}} &= \frac{1}{2} \\ 1 - \frac{v^2}{c^2} &= \frac{1}{4} \\ \frac{v^2}{c^2} &= 1 - 0.25 = 0.75 \\ v &= c\sqrt{0.75} \\ v &= 3 \times 10^8 \times 0.866 \\ v &= 2.598 \times 10^8 \text{ m/s} \end{aligned} $$ The energy required to reach this speed is equal to the relativistic kinetic energy $K$: $$ \begin{aligned} K &= (m - m_0)c^2 \\ K &= (2m_0 - m_0)c^2 \\ K &= m_0 c^2 \\ K &= 1.67 \times 10^{-27} \times (3 \times 10^8)^2 \\ K &= 1.67 \times 10^{-27} \times 9 \times 10^{16} \\ K &= 1.503 \times 10^{-10} \text{ J} \end{aligned} $$ Thus, the required speed is $2.598 \times 10^8 \text{ m/s}$ and the required energy is $1.503 \times 10^{-10} \text{ J}$.