Let's see:
250 rounds -> 1 min = 60 sec
1/4 rounds => 60/(4*250) sec =
0.06 secu1=250 rpm =>
w1 = 25*pi/3 rad/secu2=50 rpm =>
w2 = 5*pi/3 rad/secThe difference in the kinetic energies equals the torque work:
dKE = W =>
1/2*I*w1^2 - 1/2*I*w2^2 = F*r*angle =>
1/2*I * (w1^2 - w2^2) = F*r*angle = T*angle =>
T = 1/2*
I * (w1^2 - w2^2) / angle = 1/2*
(m*r^2 + I_rotor) * (w1^2 - w2^2) / angle =>
T = 1/2*
(0.026*0.02^2 + 0) * [ (25^2*pi^2-5^2*pi^2) / 3^2 ] / (pi/2) =>
T = 0.0021781709065 N*m 
I am a little bit rusty at physics and calculations but perhaps I got it right this time.
Anyway, a motor with an average volume of 16 cm^3 that can give a torque greater that this, will be useful too.
In my calculations I assume that
the inertia of the rotor is zero and that
the motor responds and accelerates instantaneously.
Generally, I don't think any of these are true in a real application so it good to demand a greater torque than this.
I don't think that a motor at this scale will be such a great cost, so perhaps with a more powerful motor I could even increase the average rotation speed (I don't really have an upper limit on this value). The ability to increase the weight if needed would be nice too.
Perhaps we should get it the other way around : from DC motors with an average motor-volume of 16 cm^3, what's the maximum torque (without sacrificing the rotation speed) that I can get and which motor provides it?
EDIT : This site does not seem to support unicode so I had to change some unicode letters that I had used.
EDIT2 : Added some more info.