Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The center frequency is given by
$$ f_0 = \frac{c}{\lambda_0} $$
where $ c $ is the speed of light and the $ \lambda_0 $ is the wavelength.
Plugging the known:
$$ f_0 = \frac{3.0 \times 10^8}{600 \times 10^{-9}} =\color{red}{\bf 5.0 \times 10^{14} }\; \text{Hz} $$
$$\color{blue}{\bf [b]}$$
The number of cycles $ N $ is given by
$$ N =\dfrac{\Delta t}{T}= f_0 \cdot \Delta t $$
where $ \Delta t $ is the pulse duration and $T$ is the time period of the center frequency where $T=1/f_0$
Plugging the known:
$$ N = (5.0 \times 10^{14} ) (6.0 \times 10^{-15} ) = \color{red}{\bf 3.0} \; \text{cycles} $$
$$\color{blue}{\bf [c]}$$
The relationship between the pulse duration $ \Delta t $ and the frequency range $ \Delta f $ (bandwidth) is given by t
$$ \Delta f \Delta t \approx 1 $$
Hence,
$$\Delta f\approx \frac{1}{\Delta t}$$
Plug the known;
$$ \Delta f = \frac{1}{\Delta t} = \frac{1}{6.0 \times 10^{-15} } = \bf 1.67 \times 10^{14} \; \text{Hz} $$
Thus, the range of frequencies $ \Delta f $ is approximately $ 1.67 \times 10^{14} \; \text{Hz} $.
The pulse consists of frequencies in the range:
$$ f_0 - \frac{\Delta f}{2} \leq f \leq f_0 + \frac{\Delta f}{2} $$
So the range of frequencies is:
$$ (5.0 \times 10^{14} - 0.835 \times 10^{14}) \leq f \leq (5.0 \times 10^{14} + 0.835 \times 10^{14}) $$
$$\color{red}{\bf 4.165 \times 10^{14} }\; \text{Hz} \leq f \leq \color{red}{\bf 5.835 \times 10^{14} }\; \text{Hz} $$
$$\color{blue}{\bf [d]}$$
The spatial length $ L $ of the pulse is the distance the light travels during the time $ \Delta t $, so
$$ L = c \Delta t $$
Plugging the known;
$$ L = (3.0 \times 10^8 ) \times (6.0 \times 10^{-15} ) = 1.8 \times 10^{-6} \; \text{m} = \color{red}{\bf 1.8} \; \mu\text{m} $$
Thus, the spatial length of the laser pulse is **1.8 micrometers (µm)**.
$$\color{blue}{\bf [e]}$$
See the graph below.
