Apply Kirchhoff's Voltage Law for a series circuit.

In a series circuit, the total voltage supplied by the battery is the sum of the voltage drops across each resistor.

$$V = V_1 + V_2 + V_3$$

Substitute the given values from Figure 1b:

$$25\,V = 10\,V + 4\,V + V_3$$ $$25\,V = 14\,V + V_3$$ $$V_3 = 25\,V - 14\,V$$ $$V_3 = \boxed{\text{11 V}}$$

Determine the current flowing through resistor R1

The total current $I$ flowing through a series circuit is determined by the total voltage $V$ and the total resistance $R_{total}$. In a series circuit, the current is the same through all resistors, including $R_1$.

$$I = \frac{V}{R_{total}}$$

Substitute the total voltage $V = 25\,V$ and the given total resistance $R_{total} = 100\,\Omega$:

$$I = \frac{25\,V}{100\,\Omega}$$ $$I = \boxed{\text{0.25 A}}$$

Find also the value of resistor R2

The resistance of $R_2$ can be found using the voltage drop across it ($V_2$) and the current flowing through it ($I$).

$$R_2 = \frac{V_2}{I}$$

Substitute the given $V_2 = 4\,V$ from Figure 1b and the calculated current $I = 0.25\,A$:

$$R_2 = \frac{4\,V}{0.25\,A}$$ $$R_2 = \boxed{\text{16 }\Omega}$$