Answer
See the detailed answer below.
Work Step by Step
We know that the relativistic kinetic energy is given by
$$K=(\gamma_p-1 )mc^2$$
where $v=0.8c$ is the speed of the particle, and $\gamma_p=\left[1-\frac{v^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.8^2\color{red}{\bf\not}c^2}{\color{red}{\bf\not}c^2}\right]^{-1/2}=\bf \frac{5}{3}$
Thus,
$$K=\left( \dfrac{5}{3}-1 \right)mc^2= \frac{2}{3} mc^2$$
Plug the known;
$$K= \frac{2}{3} (1\times 10^{-3})(3\times 10^8)^2 $$
$$K= \color{red}{\bf 6.0\times 10^{13}}\;\rm J$$
Recalling that the rest energy is given by
$$E_0=mc^2=(1\times 10^{-3})(3\times 10^8)^2$$
$$E_0= \color{red}{\bf 9.0\times 10^{13}}\;\rm J$$
The total energy is given by
$$E=\gamma_pmc^2=E_0+K$$
Plug from above;
$$E= (9.0\times 10^{13})+(6.0\times 10^{13})$$
$$E= \color{red}{\bf 1.5\times 10^{14}}\;\rm J$$
