Let the equations of two circles $$C_1$$ and $$C_2$$ be given by $$x^2 + y^2 - 4x - 4y + 6 = 0$$ and $$x^2 + y^2 - 10x - 10y + k = 0$$ respectively, where $$k$$ is a constant. Suppose that $$C_1$$ and $$C_2$$ have exactly two common tangents. Then possible values of $$k$$ are

Centre and radius of a circle $$x^2 + y^2 +2gx +2fy + c = 0$$ is (-g,-h) and $$\sqrt{g^{2}+h^{2}-c}$$
Centre and the radius of $$C_1$$ is (2,2) and $$\sqrt{2}$$
Centre and the radius of $$C_2$$ is (5,5) and $$\sqrt{50-k}$$
Distance between the centres of the two circles = $$\sqrt{18}$$
According to the properties of the circle 
If two circles have two common tangents then ,
|Difference of the radii|  < Distance between the centres of the two circles < |Sum of the radii|
$$\sqrt{50-k}$$ - $$\sqrt{2}$$ <  $$\sqrt{18}$$ < $$\sqrt{50-k}$$ + $$\sqrt{2}$$
Let's look at the options one by one ,

Option A : let k= 1  |7-$$\sqrt{2}$$| < 3$$\sqrt{2}$$ < |7+$$\sqrt{2}$$|  ------------------->  Does not satisfy 
Option B : let k= 25 |5-$$\sqrt{2}$$| < 3$$\sqrt{2}$$ < |5+$$\sqrt{2}$$|  -------------------> Satisfied
Option C : let k= 49  |1-$$\sqrt{2}$$| < 3$$\sqrt{2}$$ < |1+$$\sqrt{2}$$|  ------------------->  Does not satisfy  

Hence B is the correct answer.