Answer
9600
$3.5\times10^{-15}s$
Work Step by Step
The length of the pulse is $d = c\Delta t$, where the time is 34 ps.
Find the number of wavelengths that fit within this distance.
Number of wavelengths that fit = $\frac{c\Delta t}{\lambda}=\frac{(3.00\times10^8m/s)( 34\times10^{-12}s)}{1062\times10^{-9}m}=9600$
If the original 34 picosecond pulse were 9600 times shorter, it would contain only one wavelength. Find out how brief such a pulse would have to be.
$$\Delta tâ=\frac{34\times10^{-12}s }{9600}=3.5\times10^{-15}s$$