If ##\tau= 0.0727, N=60, i=1.3, B=1.0,## and ##\theta=15##, I tried the following calculation:
##\tau=NIABsin\theta##
##\tau=NIs^2Bsin\theta##
##s^2=\frac {\tau} {NIBsin\theta}=\frac {.0727} {60*1.3*1*sin(15)}=0.0632 m=6.32 cm##
The answer is probably right in front of me, but I don't know what...