We have the relation
[tex]\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}[/tex]
where [tex]v_{A \mid B}[/tex] denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.
We're given speeds
[tex]v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}[/tex]
[tex]v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}[/tex]
Let's assume the river flows South-to-North, so that
[tex]\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath[/tex]
and let [tex]-90^\circ < \theta < 90^\circ[/tex] be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that
[tex]\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)[/tex]
Then the velocity of the boat relative to the Earth is
[tex]\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath[/tex]
The crossing is 153.0 m wide, so that for some time [tex]t[/tex] we have
[tex]153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s[/tex]
which is minimized when [tex]\theta=0^\circ[/tex] so the crossing takes the minimum 30.0 s when the boat is pointing due East.
It follows that
[tex]\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}[/tex]
The boat's position [tex]\vec x[/tex] at time [tex]t[/tex] is
[tex]\vec x = \vec v_{B\mid E} t[/tex]
so that after 30.0 s, the boat's final position on the other side of the river is
[tex]\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath[/tex]
and the boat would have traveled a total distance of
[tex]\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}[/tex]
