Answer
a) $4.86$
b) $1.458\times 10^9\;\rm K$
Work Step by Step
a) The first reaction of the carbon cycle is given by
$$ \rm ^{12}_6C+^2_1H\rightarrow\;\;^{13}_7N+\gamma$$
In example 31-9, we can see that the energy needed is given by
$$KE=\dfrac{kQ_{ \rm ^{12}_6C}Q_{\rm ^2_1H}}{r_{ \rm ^{12}_6C}+r_{\rm ^2_1H}}$$
Thus;
$$KE=\dfrac{k (6e)(e)}{r_{ \rm ^{12}_6C}+r_{\rm ^2_1H}}$$
$$KE=\dfrac{6k e^2}{r_{ \rm ^{12}_6C}+r_{\rm ^2_1H}}$$
We know that the radius of a nucleus is given by
$$r=(1.2\times 10^{-15})A^{\frac{1}{3}}$$
Thus,
$$KE=\dfrac{6k e^2}{(1.2\times 10^{-15}) \left[ A_{\rm C}^{\frac{1}{3}}+A_{\rm H}^{\frac{1}{3}} \right]}$$
Plugging the known;
$$KE=\dfrac{6(8.99\times 10^9)(1.6\times 10^{-19})^2}{(1.2\times 10^{-15})\left[ 12^{\frac{1}{3}}+1^{\frac{1}{3}} \right]}$$
$$KE=3.498\times 10^{-13}\;\rm\color{red}{\bf\not} J\cdot \dfrac{\rm MeV}{10^6\times 1.6\times 10^{-19}\;\rm\color{red}{\bf\not} J}$$
$$KE=\bf 2.1864\;\rm MeV$$
The ratio needed is given by
$$\dfrac{KE_{\rm C}}{KE_{\rm d-t}}=\dfrac{2.1864}{0.45}=\color{red} {\bf 4.86}$$
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b) We know that kinetic energy is proportional to temperature.
$$KE\propto T$$
Thus,
$$\dfrac{KE_{\rm C}}{KE_{\rm d-t}}=\dfrac{T_{\rm C}}{T_{\rm d-t}}=4.86$$
Solving for $T_{\rm C}$;
$$T_{\rm C}=4.86T_{\rm d-t}$$
Plugging the known;
$$T_{\rm C}=4.86 \left(3\times 10^8\right)=\color{red} {\bf 1.458\times 10^9}\;\rm K$$