Answer
a) $13.9\;\rm MeV$
b) $4.315\;\rm MeV$
c) $3.335\times 10^{10}\;\rm K$
Work Step by Step
a) The energy released is given by
$$2\;{\rm \left(^{12}_6C\right)\rightarrow \;^{24}_{12}Mg}+Q$$
where $Q$ is the energy released.
$$Q= 2Q_{\rm ^{12}_6C}-Q_{\rm ^{24}_{12}Mg} $$
$$Q=\left(2m_{\rm ^{12}_6C}-m_{\rm ^{24}_{12}Mg}\right)c^2$$
Plugging the known;
$$Q=\left(2[12.000000]-23.985042\right)\times 931.5$$
$$Q=\color{red}{\bf 13.9}\;\rm MeV$$
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b) We can say that the potential energy between the two nuclei will be completely converted to kinetic energy and vice versa.
Thus,
$$PE_{sys}=KE_{sys}$$
$$PE_{sys}=2KE_{\rm _6^{12}C}$$
Hence,
$$KE_{\rm _6^{12}C}=\dfrac{PE_{sys}}{2}=\dfrac{kq_1q_2}{2r}$$
$$KE_{\rm _6^{12}C} =\dfrac{k(6e)(6e)}{2r}$$
$$KE_{\rm _6^{12}C} =\dfrac{k(6e)^2}{2r}$$
Plugging the known;
$$KE_{\rm _6^{12}C} =\dfrac{8.99\times 10^9(6\times 1.6\times 10^{-19})^2}{2(6\times 10^{-15})}\times \dfrac{1}{10^6\times 1.6\times 10^{-19}}$$
the last term is to convert from J to MeV.
$$KE_{\rm _6^{12}C} =\color{red}{\bf 4.315}\;\rm MeV$$
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c) We know that
$$KE=\frac{3}{2}kT$$
Solving for $T$;
$$T=\dfrac{2KE}{3k}$$
Plugging from above and recalling to convert the kinetic energy into joules.
$$T=\dfrac{2\times 4.315\times 10^6\times 1.6\times 10^{-19}}{3\times 1.38\times 10^{-23}}$$
$$T=\color{red}{\bf3.335\times 10^{10}}\;\rm K$$