For the stationary wave $\mathrm{y}=4\sin \left(\frac{\mathrm{\pi x}}{15}\right)\cos \left(96\mathrm{\pi t}\right)$, the distance between a node and the next antinode is (in m)

Comparing given equation with standard equation 
$\mathrm{y}=2\mathrm{asin}\frac{2\mathrm{\pi x}}{\mathrm{\lambda }}\cos \frac{2\mathrm{\pi vt}}{\mathrm{\lambda }}$  gives us  $\frac{2\mathrm{\pi }}{\mathrm{\lambda }}=\frac{\mathrm{\pi }}{15}\Rightarrow \mathrm{\lambda }=30$
Distance between nearest node and antinodes = $\frac{\mathrm{\lambda }}{4}=\frac{30}{4}=7.5$