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**Hint:**To answer this question, we first need to understand what is ohm’s law. The relationship between current, voltage, and resistance is described by Ohm's law. The potential difference, or voltage, across a large number of materials is precisely proportional to the amount of continuous current flowing through them.

**Complete step by step answer:**

Voltage distribution law for series resistance - Resistors in series carry the same current, but their varied resistance values because various voltage drops across each resistor, according to Ohm's Law ($V = IR$). Then there are voltage dividers, which are series circuits.

**Case 1:**$R = 16\,Ω$ (given)

${V_R}$= 12 V

Let us assume that the internal resistance of the battery be ${R_b}$ Ω and ${V_b}$ be the voltage of the battery. So, by ohm’s law

${V_R} = \dfrac{R}{{R + {R_b}}}{V_b}$ (by voltage distribution law).

Substituting values

$12 = \dfrac{{16}}{{16 + {R_b}}}{V_b}$

By solving this equation, we get

${V_b} = \dfrac{{48 + 3{R_b}}}{4}$ (mark this as equation 1)

**Case 2:**$R = 10\,Ω$ (given)

${V_R}$= 11 V

Let us assume that the internal resistance of the battery be ${R_b}$ Ω and ${V_b}$be the voltage of the battery. So, by ohm’s law

${V_R} = \dfrac{R}{{R + {R_b}}}{V_b}$ (by voltage distribution law).

Substituting values

$11 = \dfrac{{10}}{{10 + {R_b}}}{V_b}$

By solving this equation, we get

${V_b} = \dfrac{{110 + 11{R_b}}}{{10}}$ (mark this as equation 2)

Equating the marked equation 1 and 2.

$\dfrac{{48 + 3{R_b}}}{4} = \dfrac{{110 + 11{R_b}}}{{10}}$

Solving further we get,

$240 + 15{R_b} = 220 + 22{R_b}$

We get

$20 = 7{R_b}$

$\therefore {R_b}=\dfrac{20}{7}$

So, the final value of internal resistance is 20/7.

**Therefore option B is the correct answer.**

**Note:**A battery's internal resistance (IR) is defined as the resistance to current flow within the battery. Electronic resistance and ionic resistance are the two main factors that influence a battery's internal resistance.

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