1. Initial charges on the spheres

If the initial surface charge density on both spheres is $\sigma$, then:

• For the smaller sphere of radius $R$: $q_1 = \sigma \cdot 4\pi R^2$

• For the bigger sphere of radius $3R$: $q_2 = \sigma \cdot 4\pi (3R)^2 = \sigma \cdot 4\pi \cdot 9R^2 = 9\sigma \cdot 4\pi R^2$

So total charge is $Q = q_1 + q_2 = \sigma 4\pi R^2 + 9\sigma 4\pi R^2 = 10\sigma 4\pi R^2$

2. Condition after contact

When two conducting spheres are brought into contact, their potentials become equal.

Let final charges be $Q_1$ on radius $R$ sphere and $Q_2$ on radius $3R$ sphere.

Since potential of a conducting sphere is $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r},$

equal potentials give $\frac{Q_1}{R} = \frac{Q_2}{3R}$ $Q_2 = 3Q_1$

3. Use conservation of charge

$Q_1 + Q_2 = 10\sigma 4\pi R^2$ Substitute $Q_2 = 3Q_1$: $Q_1 + 3Q_1 = 10\sigma 4\pi R^2$ $4Q_1 = 10\sigma 4\pi R^2$ $Q_1 = \frac{10}{4}\sigma 4\pi R^2 = \frac{5}{2}\sigma 4\pi R^2$

Then $Q_2 = 3Q_1 = \frac{15}{2}\sigma 4\pi R^2$

4. Find final surface charge densities

• Smaller sphere: $\sigma_1 = \frac{Q_1}{4\pi R^2} = \frac{\frac{5}{2}\sigma 4\pi R^2}{4\pi R^2} = \frac{5\sigma}{2}$

• Bigger sphere: $\sigma_2 = \frac{Q_2}{4\pi (3R)^2} = \frac{Q_2}{36\pi R^2}$

Now, $Q_2 = \frac{15}{2}\sigma 4\pi R^2 = 30\pi \sigma R^2$ So, $\sigma_2 = \frac{30\pi \sigma R^2}{36\pi R^2} = \frac{5\sigma}{6}$

5. Required ratio

$\frac{\sigma_1}{\sigma_2} = \frac{\frac{5\sigma}{2}}{\frac{5\sigma}{6}} = \frac{6}{2} = 3$

Therefore, the correct option is $\boxed{3}$ which corresponds to Option D.