Answer
$16.11N.m$
Work Step by Step
We know that
$M=\frac{2}{3}(\frac{R_2^3-R_1^3}{R_2^2-R_1^2})$
$\implies M_A=\frac{2}{3}(0.20)(3000)[\frac{(0.03)^3-(0.01)^3}{(0.03)^2-(0.01)^2}]=13.00N.m$
Similarly
$M_B=\frac{2}{3}(0.20)(1000)[\frac{(0.02)^3-(0.01)^3}{(0.02)^2-(0.01)^2}]=3.11N.m$
We also know that the sum of the moments about the z-axis is equal to zero
$\implies \Sigma M_z=0$
$\implies M-M_A+M_B=0$
$\implies M=M_A+M_B$
$\implies M=13.00+3.11$
We plug in the known values to obtain:
$M=16.11N.m$
