Take east as the positive $x$-direction and north (across the river) as the positive $y$-direction.

Velocity of river flow: $\vec v_r = 9\,\hat i\ \text{km h}^{-1}$

Velocity of boat relative to water: The boat moves with maximum speed $27\,\text{km h}^{-1}$ at an angle $150^\circ$ to the direction of river flow.

Since river flow is along east, the boat's velocity relative to water makes angle $150^\circ$ with east.

So its components are $v_{bx}=27\cos150^\circ=27\left(-\frac{\sqrt3}{2}\right)=-\frac{27\sqrt3}{2}$ $v_{by}=27\sin150^\circ=27\left(\frac12\right)=13.5$ Thus, $\vec v_{b/w}=-\frac{27\sqrt3}{2}\hat i+13.5\hat j$

Resultant velocity of boat relative to ground: $\vec v = \vec v_{b/w}+\vec v_r$ $v_x=9-\frac{27\sqrt3}{2},\qquad v_y=13.5$

For crossing the river, only the perpendicular component matters: $v_y=13.5\,\text{km h}^{-1}$

Time of crossing: Half a minute means $t=\frac{1}{2}\text{ min}=30\text{ s}=\frac{1}{120}\text{ h}$

Width of river: $\text{Width}=v_y\,t=13.5\times\frac{1}{120}\ \text{km}$ $=0.1125\ \text{km}$ Converting to metres, $0.1125\times 1000=112.5\ \text{m}$

Check options: $\boxed{112.5\ \text{m}}$ This matches Option A.