Let the center of third circle be O and radius be r.

OA = OB = OC = r
Mid point of BC is D

DC = 1 cm, OD = 3-r

OC^2 = OD^2 + DC^2

r*r = 1 + (3-r)^2

r = 5/3 cm

Answer: C

Okay it took me quite some time to get this one. But I did.
The radius of the third circle is the radius of the circumcenter of triangle ABC. Circumcenter is the point where the perpendicular bisectors of all three sides meet.

Taking A as the origin, I got B as (3,1) and C as (3, -1). The perpendicular bisector of BC is the x axis (y = 0).
So now I just need to find the intersection of y= 0 and the perpendicular bisector of AB or AC.
Equation of AB is y = mx + c
Since A is the origin, the y intercept c = 0. using B(3.1) I get the slope as 1/3.
Slope of perpendicular bisector of AB is -3 and midpoint of AB (1.5, 0.5) is a point on this line.
Equation of perpendicular bisector of AB is y - 0.5 = -3 (x - 1.5)
y = -3x + 5
Now solving y= 0 with this equation you get x = 5/3 which gives us the circumcenter (5/3, 0)
And radius of the third circle is distance from A (0,0) to the circumcenter (5/3,0) which is 5/3

I wouldn't attempt this in the exam. Takes too much time.

@nmerchant - your approach is not advisable on real GMAT. Please look my post, above.

Genius! Very efficient approach for the GMAT thanks a ton.

OAC

Hello - can someone draw a diagram for me? I'm having diffuculty conceptualizing the 3 circles and a diagram would help me figure out where I'm going wrong.

Thanks!