Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that
$$ \Delta f \Delta t \approx 1 \tag 1$$
for a wave packet, which represents the time-frequency uncertainty principle.
For a photon, the energy $ E $ and frequency $ f $ are related by:
$$ E = h f $$
Thus,
$$ \Delta E = h \Delta f $$
Solving for $\Delta f$;
$$\Delta f=\dfrac{ \Delta E}{h}$$
Plugging into (1);
$$ \dfrac{ \Delta E}{h}\Delta t \approx 1 $$
$$ \boxed{\Delta E \Delta t \approx h }$$
$$\color{blue}{\bf [b]}$$
The formula of $ \Delta E \Delta t \approx h $ tells us that there is an inherent uncertainty in the energy of the photon and the time it takes for the photon to be emitted or absorbed.
If $ \Delta t $ is large, then the uncertainty in the energy $ \Delta E $ will be small. And if $ \Delta t $ is small, then the uncertainty in the energy $ \Delta E $ will be large.
$$\color{blue}{\bf [c]}$$
The uncertainty of the photon's energy is given by the boxed formula above,
$$ \Delta E \approx \frac{h}{\Delta t} $$
Plug the known;
$$ \Delta E = \dfrac{6.626 \times 10^{-34}}{10 \times 10^{-9}} $$
$$\Delta E= 6.626 \times 10^{-26} \; \rm{J} =\color{red}{\bf 4.14 \times 10^{-7}} \; \rm{eV} $$
$$\color{blue}{\bf [d]}$$
We are given that the wavelength of the emitted photon $ \lambda = 500 \; \rm{nm} $.
And we know that the photon's energy is given by
$$ E = \frac{hc}{\lambda} $$
Plug the known;
$$ E = \frac{(6.626\times 10^{-34})(3.0 \times 10^8)}{500 \times 10^{-9}} = \bf 3.9756 \times 10^{-19} \; \rm{J} $$
The fractional uncertainty is the ratio of the uncertainty in energy to the actual energy:
$$ \dfrac{\Delta E}{E} =\dfrac{6.626 \times 10^{-26} }{3.9756 \times 10^{-19}}=\color{red}{\bf 1.667\times 10^{-7}}$$
