Question regarding supermesh
Verifying your claim of \$10\:\text{mA}\$ for the lower-left quadrant
Here's my marked-up version of your schematic:

Before I delve too much into a general solution, let's verify your assertion about \$I_3=10\:\text{mA}\$. It's not hard to do and you are right about that.
By inspection, you know that \$V_x=R_5\cdot\left(I_4-I_2\right)\$. But also by inspection, you know that \$g\cdot V_x=I_4-I_3\$. And from the problem statement, you know that \$I_2=I_s\$. So it must be that: \$g\cdot R_5\cdot\left(I_4-I_2\right)=I_4-I_3\$. Since \$g\cdot R_5=1\$, given your problem statement, this means \$I_4-I_2=I_4-I_3\quad\therefore I_3=I_2\$. And as \$I_2=I_s=10\:\text{mA}\$, you have your answer that \$I_3=10\:\text{mA}\$.
So I completely agree with your finding for the above reasoning.
General Solution
Now, we could use that fact to reduce the above problem somewhat and find the rest. But I don't want to do that. I want to take a very general approach to the above problem and solve it, generally. It's better to avoid using special-case findings, as above, because if you learn the general approaches well, you don't need to waste time finding all the nooks and crannies of a particular circuit to simplify it. Instead, you just use very solid, bullet-proof methods that will consistently grind out good results time after time after time.
So let's just grind-out the mesh equations. I will take each successive quadrant by starting out at the lower-left node of the loop, assigning that node voltage the value of \$0\:\text{V}\$, and then work around the loop to develop the equation. I'll do this for all four loops.
$$\begin{align*}
0\:\text{V}+V_s-R_1\cdot I_1-R_3\cdot\left(I_1-I_2\right)-R_2\cdot\left(I_1-I_3\right)&amp;=0\:\text{V}\\0\:\text{V}-R_3\cdot\left(I_2-I_1\right)-R_4\cdot I_2+V_{\text{I}_S}-R_5\cdot\left(I_2-I_4\right)&amp;=0\:\text{V}\\0\:\text{V}-r\cdot \left(I_x=I_4\right)-R_2\cdot\left(I_3-I_1\right)-V_{g\,V_x}&amp;=0\:\text{V}\\0\:\text{V}+V_{g\,V_x}-R_5\cdot\left(I_4-I_2\right)-R_6\cdot I_4&amp;=0\:\text{V}
\end{align*}$$
Now, we know one more thing from the earlier analysis above:
$$g\cdot R_5\cdot\left(I_4-I_2\right)=I_4-I_3$$
So this provides five equations. Since we already are given \$I_s=I_2=10\:\text{mA}\$ in the problem specifications, we have these five unknowns to solve: \$I_1\$, \$I_3\$, \$I_4\$, \$V_{\text{I}_S}\$, and \$V_{g\,V_x}\$.
Let's use sympy:
var('r1 r2 r3 r4 r5 r6 r g is i1 i2 i3 i4 vgx vis vs')
eq1=Eq(0+vs-r1*i1-r3*(i1-i2)-r2*(i1-i3),0)
eq2=Eq(0-r3*(i2-i1)-r4*i2+vis-r5*(i2-i4),0)
eq3=Eq(0-r*i4-r2*(i3-i1)-vgx,0)
eq4=Eq(0+vgx-r5*(i4-i2)-r6*i4,0)
eq5=Eq(g*r5*(i4-i2),i4-i3)
ans=solve([eq1,eq2,eq3,eq4,eq5],[i1,i3,i4,vgx,vis])
for x in ans:x,ans[x].subs({r1:1e3,r2:500,r3:1e3,r4:1e3,r5:1e3,r6:1e3,r:1e3,g:.001,vs:20,i2:10e-3})
(i3, 0.0100000000000000)
(vgx, -2.00000000000000)
(vis, 12.0000000000000)
(i1, 0.0140000000000000)
(i4, 0.00400000000000000)

Here indeed, you can see that \$I_3=10\:\text{mA}\$. We knew this, already. But this just confirms it, but from a far more general approach to the problem.
Note that there was no need for a supermesh. There never is. Just as with KCL and nodal, there never is a need for a supernode. They are always superfluous.