In a single-slit experiment, the slit width is 150 times the wavelength of the light. What is the width (in mm) of the central maximum on a screen 2.6 m behind the slit?

I have tried:

y=[(1+1/2)(lambda)(2.6m)] / (150lambda)

to bring me to

y=[(1.5)(2.6)] \ (150)

giving me a y value in mm of 26