Answer
See the detailed answer below.
Work Step by Step
a) First, we need to use the uncertainty principle, as the author told us, but let's make an assumption of $E\approx pc$ since the given energy is very big.
Thus;
$$\Delta p=\dfrac{\Delta E}{c}$$
Multiplying both sides by $\Delta x$
$$\Delta p\Delta x=\dfrac{\Delta E}{c}\Delta x$$
The left side is now equal to $h/2pi$, so
$$\dfrac{h}{2\pi}=\dfrac{\Delta E}{c}\Delta x$$
Solving for $\Delta E$;
$$\Delta E=\dfrac{hc}{2\pi \Delta x}$$
To find the result in eV, we need to divide by $e$;
$$\Delta E=\dfrac{hc}{2\pi e\Delta x}$$
Plugging the known;
$$\Delta E=\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{2\pi (1.6\times 10^{-19})(10^{-31})}$$
$$\Delta E= 1.977\times 10^{24}\;\rm eV\approx \color{red}{\bf 2.0\times 10^{15}}\;\rm GeV$$
which is close to the given value.
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Now we need to use de Broglie’s formula;
$$\lambda=\dfrac{h}{p}$$
and we know, from the assumption above, that $p=E/c$. so that
$$\lambda=\dfrac{hc}{E}$$
Solving for $E$;
$$E=\dfrac{hc}{\lambda}$$
To find the result in eV, we need to divide by $e$;
$$E=\dfrac{hc}{e\lambda}$$
Plugging the known; where $\lambda$ here is assumed to be the unification distance.
$$E=\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{(1.6\times 10^{-19})10^{-31}}$$
$$ E= 1.24237\times 10^{25}\;\rm eV\approx \color{red}{\bf 1.0\times 10^{16}}\;\rm GeV$$
which is close to the given value as well.
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b) We know that
$$E=\frac{3}{2}kT$$
Solving for $T$;
$$T=\dfrac{2E}{3k}$$
Plugging the known and do not forget to convert the energy to joules.
$$T=\dfrac{2(10^{16}\times 10^9)(1.6\times 10^{-19})}{3 (1.38\times 10^{-23})}$$
$$T=\color{red}{\bf 7.723\times 10^{28}}\;\rm K$$
