Answer
(a) $N_\text{emitted}= 9.0 \times 10^5\;\rm photons $
(b) $\tau=8.69 \rm \;ns$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since each atom emits one photon when it undergoes a quantum jump to the ground state, the number of photons emitted is the same as the number of atoms that have undergone the transition.
We know that 90% of $ 1.0 \times 10^6 $ atoms have emitted a photon, so
$$
N_\text{emitted} = 0.90 \times (1.0 \times 10^6) = \color{red}{\bf 9.0 \times 10^5}\;\rm photon
$$
$$\color{blue}{\bf [b]}$$
The decay of the number of excited atoms is given by
$$
N_{\rm excited}= N_0 \, e^{-t/\tau}
$$
where $ N_{\rm excited} $ is the number of atoms remaining excited at time $ t =20$ ns, $ N_0 $ is the initial number of excited atoms, and $ \tau $ is the lifetime of the excited state.
We know that after 20 ns, 90% of the atoms have decayed, meaning 10% remain in the excited state.
So, $ N_{\rm excited} =0.1N_0$
$$
0.1\color{red}{\bf\not} N_0 = \color{red}{\bf\not} N_0 \, e^{-20/\tau}
$$
Taking the natural logarithm on both sides:
$$
\ln(0.10) = -\frac{20}{\tau}
$$
Thus,
$$
\tau = \frac{-20}{\ln(0.10) } = \color{red}{\bf 8.69}\, \text{ns}
$$
