A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is _____ $$kg.m^{2}$$.

We have two solid spheres in contact: Sphere A (mass $$m_1 = 5$$ kg, radius $$r_1 = 0.1$$ m) and Sphere B (mass $$m_2 = 10$$ kg, radius $$r_2 = 0.2$$ m). We need the moment of inertia about a tangent at the point of contact.

Let the point of contact be T. The tangent at T is perpendicular to the line joining the centres. The centre of Sphere A is at distance $$r_1 = 0.1$$ m from T, and the centre of Sphere B is at distance $$r_2 = 0.2$$ m from T (on the other side).

For a solid sphere, the moment of inertia about its own centre is $$I_{cm} = \dfrac{2}{5}mr^2$$. Using the parallel axis theorem, the moment of inertia about a line at distance $$d$$ from the centre is $$I = I_{cm} + md^2$$.

For Sphere A (distance from centre to tangent = $$r_1$$):
$$I_A = \frac{2}{5}m_1 r_1^2 + m_1 r_1^2 = \frac{7}{5}m_1 r_1^2$$
$$I_A = \frac{7}{5} \times 5 \times (0.1)^2 = \frac{7}{5} \times 5 \times 0.01 = 7 \times 0.01 = 0.07 \text{ kg.m}^2$$

For Sphere B (distance from centre to tangent = $$r_2$$):
$$I_B = \frac{2}{5}m_2 r_2^2 + m_2 r_2^2 = \frac{7}{5}m_2 r_2^2$$
$$I_B = \frac{7}{5} \times 10 \times (0.2)^2 = \frac{7}{5} \times 10 \times 0.04 = 14 \times 0.04 = 0.56 \text{ kg.m}^2$$

Thus the total moment of inertia is
$$I = I_A + I_B = 0.07 + 0.56 = 0.63 \text{ kg.m}^2$$

The answer is $$\boxed{0.63}$$ kg.m$$^2$$, which corresponds to Option 4.