Since [tex]M[/tex] is closed, you can use the divergence theorem: The flux of [tex]\vec E(x,y,z)[/tex] across [tex]M[/tex] is
[tex]\displaystyle\iint_{\partial M}\vec E\cdot\mathrm d\vec S=\iiint_M(\nabla\cdot\vec E)\,\mathrm dV=54\iiint_M\mathrm dV[/tex]
which is 54 times the volume of the hemisphere centered at (0, 0, 0) with radius 1, [tex]\boxed{36\pi}[/tex].
Judging by the question content, you're supposed to find this value by computing the the integral of [tex]\vec E[/tex] across [tex]M_1[/tex] and [tex]M_2[/tex].
Parameterize the hemisphere by
[tex]\vec r(u,v)=(\cos u\sin v,\sin u\sin v,\cos v)[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\frac\pi2[/tex]. Take the normal vector to [tex]M_1[/tex] to be
[tex]\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=(\cos u\sin^2v,\sin u\sin^2v,\sin v\cos v)[/tex]
The flux of [tex]\vec E[/tex] across [tex]M_1[/tex] is
[tex]\displaystyle\iint_{M_1}\vec E\cdot\mathrm d\vec S=18\int_0^{\pi/2}\int_0^{2\pi}(\cos u\sin v,\sin u\sin v,\cos v)\cdot\left(\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}\right)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle18\int_0^{\pi/2}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv=36\pi[/tex]
Parameterize the disk by
[tex]\vec s(u,v)=(u\cos v,u\sin v,0)[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal to [tex]M_2[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec u}{\partial v}=(0,0,-u)[/tex]
Then the flux across [tex]M_2[/tex] is
[tex]\displaystyle\iint_{M_2}\vec E\cdot\mathrm d\vec S=18\int_0^{2\pi}\int_0^1(u\cos v,u\sin v,0)\cdot\left(\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right)\,\mathrm du\,\mathrm dv=0[/tex]
Then the total flux across [tex]M[/tex] is [tex]36\pi[/tex], as expected.
