C

Step 1: Understand the system - Two masses (each of mass 'm') are connected by a spring and one mass is given a velocity 'v = 2k', where 'k' is the spring constant.
Step 2: Conservation of momentum - Initially, mass 'm' moves with velocity 'v', therefore, the total initial momentum of the system is 'mv'. Since the other mass is at rest, the entire momentum is carried by the first mass.
Step 3: Maximum extension - At maximum extension of the spring, both masses will momentarily come to rest. This means that all kinetic energy is converted to potential energy of the spring. The initial kinetic energy is given by:
$$ KE = \frac{1}{2} mv^2 = \frac{1}{2} m(2k)^2 = 2mk^2 $$
The potential energy stored in the spring at maximum extension 'x' is given by:
$$ PE = \frac{1}{2} kx^2 $$
Setting the kinetic energy equal to potential energy:
$$ 2mk^2 = \frac{1}{2} kx^2 $$
Step 4: Solve for 'x':
$$ 2mk^2 = \frac{1}{2} kx^2 \implies 4mk^2 = kx^2 \implies x^2 = 4m \implies x = 2\sqrt{m} $$
This shows that the maximum extension 'x' is proportional to '2', confirming the relation with mass 'm'. Hence, the correct extension is given by the option in the form $2\sqrt{m}$.
Therefore, the correct answer is Option C.