1. Let the centre of the circle be $C(h,k)$.

2. Since the centre lies on the line $x+y=2$, we have $h+k=2.$

3. The circle touches the line $x=3$. So the radius equals the perpendicular distance from the centre to this line: $r=|h-3|.$

4. The circle also touches the line $y=2$. Hence, $r=|k-2|.$

5. Therefore, $|h-3|=|k-2|.$

6. The centre is in the first quadrant and also satisfies $h+k=2$, so $0<h<2$ and $0<k<2$. Thus,

   \qquad |k-2|=2-k.$$ Hence, $$3-h=2-k.$$

7. Simplify: $k-h=1.$

8. Now solve the system $h+k=2,$ $k-h=1.$

   Adding, $2k=3 \implies k=\frac{3}{2}.$ Then $h=2-\frac{3}{2}=\frac{1}{2}.$

9. Radius: $r=3-h=3-\frac{1}{2}=\frac{5}{2}.$ (Also $r=2-k=2-\frac{3}{2}=\frac{1}{2}$ seems inconsistent, so let us check carefully.)

10. Recheck the distance to the line $x=3$: Distance from $(h,k)$ to $x=3$ is $|h-3|=3-\frac{1}{2}=\frac{5}{2}.$ Distance to the line $y=2$ is $|k-2|=2-\frac{3}{2}=\frac{1}{2}.$ These are not equal, so such a circle cannot touch both lines if centre is taken strictly with $0<h<2,0<k<2$ using that branch.

11. Instead, touching both lines means the centre must be equidistant from the two lines: $|h-3|=|k-2|.$ Since the centre is in first quadrant and on $h+k=2$, let $k=2-h$. Then $|h-3|=|2-h-2|=|-h|=h,$ because $h>0$.

Also, since $h<2<3$, we have $|h-3|=3-h.$

Therefore, $3-h=h \implies 2h=3 \implies h=\frac{3}{2}.$ Then $k=2-\frac{3}{2}=\frac{1}{2}.$

12. So the radius is $r=|h-3|=3-\frac{3}{2}=\frac{3}{2}.$

13. Therefore the diameter is $2r=2\cdot \frac{3}{2}=3.$

Hence, the required diameter is $3$.