Let the internal resistance of each cell be $$r$$.

Since the two cells are connected in series, their internal resistances add up, and the total internal resistance of the series combination is $$2r$$. The total EMF of the series combination of cells is $$1.5\,\text{V} + 1.5\,\text{V} = 3\,\text{V}$$.

Let's use Kirchhoff's Voltage Law (KVL) for the closed loop in the circuit:

$$\text{EMF}_{total} - I(R + 2r) = 0$$

We are given that the voltage across the $$10\,\Omega$$ resistor, as measured by the ideal voltmeter, is $$1.5\,\text{V}$$. According to Ohm's law, the current in the circuit can be determined as:

$$I = \frac{V}{R} = \frac{1.5\,\text{V}}{10\,\Omega} = 0.15\,\text{A}$$

Now, substitute the given values into the KVL equation:

$$3\,\text{V} - 0.15\,\text{A}(10\,\Omega + 2r) = 0$$

Solve for $$2r$$:

$$3\,\text{V} - 1.5\,\text{V} = 0.15\,\text{A} \cdot 2r$$

$$1.5\,\text{V} = 0.3\,\text{A} \cdot r$$

Now, solve for the internal resistance $$r$$:

$$r = \frac{1.5\,\text{V}}{0.3\,\text{A}} = 5\,\Omega$$

So, the internal resistance of each cell is $$5\,\Omega$$.