Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the intensity of a laser beam is given by
$$I=\dfrac{P}{A } $$
So to increase $I$ by 4 times, we need to decrease the cross-sectional area $A$ of the beam by 4 times as well. Note that the power of the beam is unchangeable.
Hence,
$$I'=\dfrac{P}{\frac{1}{4}A } =4I$$
This happens when the radius of the beam is reduced to $r'=\dfrac{r}{2}$
From the geometry of the figure below,
$$\dfrac{f}{r}=\dfrac{x'}{r'}=\dfrac{x'}{\frac{1}{2}r}$$
Hence,
$$\boxed{x'=\frac{1}{2}f}$$
This is at the middle between the center of the lens and its focal point.
$$\color{blue}{\bf [b]}$$
We know that the intensity of a laser beam is also given by
$$I =\dfrac{c\epsilon_0 E_0^2}{2}$$
So to increase the electric field by 4 times, you need to increase the intensity 16 times. Note the square of $E_0$.
This means that we need to reduce the area 16 times.
And to do so, we need to decrease the radius by $1/4$.
From the geometry of the figure below,
$$\dfrac{f}{r}=\dfrac{x''}{r''}=\dfrac{x''}{\frac{1}{4}r}$$
Hence,
$$\boxed{x''=\frac{1}{4}f}$$
This is at a distance of $\frac{3}{4}f$ from the center of the lens.
