Answer
Please see the work below.
Work Step by Step
We know that
$\lambda=\frac{v}{f}$
$\implies \lambda=\frac{343m/s}{185 Hz}$
$\lambda=1.85m$
Now the distance from source 1 to student B is given as
$l_2=\sqrt{(3.0)^2+(1.5)^2}$
$l_2=3.35m$
The distance from source 1 to student is A is
$l_1=1.5$
$l_2-l_1=3.35-1.5=1.85m$
The ratio of path difference to the wavelength is given as $\frac{l_2-l_1}{\lambda}=\frac{1.85m}{1.85m}=1$
Since the difference is 1 wavelength, the tone heard at location B is minimum.
