Answer
$237.34N.m$
Work Step by Step
We know that
$M=\frac{2}{3}\mu P(\frac{R_2^3-R_1^3}{R_2^2-R_1^2})$
Now $M_A=\frac{2}{3}(0.20)(12000)[\frac{(0.10)^3-(0.05)^3}{(0.10)^2-(0.05)^2}]=186.67N.m$
and $M_B=\frac{2}{3}(0.20)(4000)[\frac{(0.075)^3-(0.05)^3}{(0.075)^2-(0.05)^2}]=50.67N.m$
We equate the moments about the z-axis equal to zero
$\implies \Sigma M_z=0$
$\implies M-M_A-M_B=0$
$\implies M=M_A+M_B$
We plug in the known values to obtain:
$M=186.67+50.67=237.34N.m$
