Answer
the gap is $3m$.
Work Step by Step
We have the information about the car ($m_{car}=1500kg$ and $l_{car}=3m$) and about the boat ($m_{boat}=4000kg$ and $l_{boat}=14m$).
The car will move on the boat from the far end to the end near the dock. While doing so, the system of the boat and the car receive no outside forces, so the system Center of Mass will not change.
The initial position of the Center of Mass (dock position as zero) is as follows:
$x_{cm}=\frac{x_{boat}m_{boat}+x_{car}m_{car}}{m_{boat}+m_{car}}$
with $x_{boat} = \frac{1}{2} l_{boat}=7m$ and $x_{car} = l_{boat}-\frac{1}{2} l_{car}=12.5m$, we obtain
$x_{cm} = 8.5m$
When the car arrives at the boat end near the dock, the boat has moved slightly, $\Delta x$, so the Center of Mass equation is:
$x_{cm}=\frac{x_{boat}m_{boat}+x_{car}m_{car}}{m_{boat}+m_{car}}$
with $x_{boat}=\Delta x+7m$ and $x_{car}=\Delta x+1.5m$, we obtain
$x_{cm}=\frac{(\Delta x+7m)m_{boat}+(\Delta x+1.5m)m_{car}}{m_{boat}+m_{car}}$
$8.5m=\Delta x+5.5m$
$\Delta x = 3m$
So, when the car is about to make a jump, the gap is $3m$.
