Let $${R_1},{R_2}$$  and $${R_3}$$ are the resistances of three bulbs respectively.
In series order, $$R = {R_1} + {R_2} + {R_3}$$
but, $$R = \frac{{{V^2}}}{P}$$  and supply voltage in series order is the same as the rated voltage.
$$\eqalign{ & \therefore \frac{{{V^2}}}{P} = \frac{{{V^2}}}{{{P_1}}} + \frac{{{V^2}}}{{{P_2}}} + \frac{{{V^2}}}{{{P_3}}} \cr & {\text{or}}\,\,\frac{1}{P} = \frac{1}{{60}} + \frac{1}{{60}} + \frac{1}{{60}} \cr & {\text{or}}\,\,P = \frac{{60}}{3} = 20\,W \cr} $$
Alternative
As three bulbs have same power and voltage so, they have equal resistance. So, power equivalent when connected in series is given by,
$$\eqalign{ & \frac{1}{{{P_{EQ}}}} = \frac{1}{{{P_1}}} + \frac{1}{{{P_2}}} + \frac{1}{{{P_3}}} \cr & \frac{1}{{{P_{EQ}}}} = \frac{1}{P} + \frac{1}{P} + \frac{1}{P}\,\,\left( {{\text{as}}\,\,{P_1} = {P_2} = {P_3} = P} \right) \cr & {\text{So,}}\,\,{P_{EQ}} = \frac{P}{3} \Rightarrow \frac{{60}}{3} = 20\,W \cr} $$

Let $${i_a}$$ be the current flowing through ammeter and $$i$$ be the total current. So, a current $$i - {i_a}$$  will flow through shunt resistance.
Potential difference across ammeter and shunt resistance is same.
i.e. $${i_a} \times R = \left( {i - {i_a}} \right) \times S$$

or $$S = \frac{{{i_a}R}}{{i - {i_a}}}\,.......\left( {\text{i}} \right)$$
Given, $${i_a} = 100\,A,i = 750\,A,R = 13\,\Omega $$
Hence, $$S = \frac{{100 \times 13}}{{750 - 100}} = 2\,\Omega $$

In potentiometer experiment in which we find internal resistance of a cell. Let $$E$$ be the emf of the cell and $$V$$ be the terminal potential difference, then
$$\frac{E}{V} = \frac{{{l_1}}}{{{l_2}}}$$
where, $${{l_1}}$$ and $${{l_2}}$$ are lengths of potentiometer wire with and without being short circuited through a resistance.
Since, $$\frac{E}{V} = \frac{{R + r}}{R}\,\,\left[ {\because E = I\left( {R + r} \right)\,{\text{and}}\,V = IR} \right]$$
$$\eqalign{ & \therefore \frac{{R + r}}{R} = \frac{{{l_1}}}{{{l_2}}} \cr & {\text{or}}\,\,1 + \frac{r}{R} = \frac{{110}}{{100}} \cr & {\text{or}}\,\,\frac{r}{R} = \frac{{110}}{{100}} \cr & {\text{or}}\,\,r = \frac{1}{{10}} \times 10 \cr & = 1\,\Omega \cr} $$

According to maximum power transfer theorem, the power output across load due to a cell or battery is maximum, if the load resistance is equal to the internal resistance of cell or battery. Hence, value of $$R$$ will be $$0.5\,\Omega .$$

From the given figure, the resistances in the arms $$AC$$ and $$BC$$ are in series.
$$\eqalign{ & \therefore R' = {R_{AC}} + {R_{BC}} \cr & = 3\,\Omega + 3\,\Omega \cr & = 6\,\Omega \cr} $$
Now, $${R'}$$ is in parallel with the resistance in arm $$AB,$$
So, $$R = \frac{{R' \times {R_{AB}}}}{{R' + {R_{AB}}}} = \frac{{6\,\Omega \times 3\,\Omega }}{{6\,\Omega + 3\,\Omega }} = 2\,\Omega $$
Therefore, the Ohm's law can be stated as
$$V = iR\,\,{\text{or}}\,\,{\text{current,}}\,i = \frac{V}{R}$$
Substituting the values, we get
$$i = \frac{2}{2} = 1\,A\,\,\left( {\because V = 2\,V} \right)$$

The balancing length for $$R$$ (when 1, 2 are connected) be $${l_1}$$ and balancing length for $$R + X$$  (when 1, 3 is connected) is $${l_2}.$$
$$\eqalign{ & {\text{Then,}}\,\,iR = k{l_1}\,{\text{and}}\,i\left( {R + X} \right) = k{l_2} \cr & {\text{Given,}}\,\,i = 1A \cr & \therefore R = k{l_1}\,.......\left( {\text{i}} \right) \cr & R + X = k{l_2}\,.......\left( {{\text{ii}}} \right) \cr & {\text{Also, subtracting Eq}}{\text{.}}\,\left( {\text{i}} \right){\text{ fom Eq}}{\text{. }}\left( {{\text{ii}}} \right){\text{, we get}} \cr & X = k\left( {{l_2} - {l_1}} \right) \cr} $$

The internal resistance of the cell,
$$r = \left( {\frac{{{\ell _1} - {\ell _2}}}{{{\ell _2}}}} \right) \times R = \frac{{240 \times 120}}{{120}} \times 2 = 2\Omega $$

$${E_{{\text{net}}}} = \frac{{{E_1}{r_2} - {E_2}{r_1}}}{{{r_1} + {r_2}}}\,or\,{E_{{\text{net}}}} = \frac{{2 - 2}}{{2 + 1}} = 0$$

We know that $$R = \frac{{V_{rated}^2}}{{{P_{rated}}}} = \frac{{{{\left( {220} \right)}^2}}}{{1000}}$$
When this bulb is connected to $$110\,volt$$  mains supply we get
$$P = \frac{{{V^2}}}{R} = \frac{{{{\left( {110} \right)}^2} \times 1000}}{{{{\left( {220} \right)}^2}}} = \frac{{1000}}{4} = 250W$$

To receive maximum current, the two batteries should be connected in series.
Given, $$R = 1\,\Omega + 1\,\Omega = 2\,\Omega .$$
Hence, power developed across the resistance $$R$$ with the batteries in series is
$$\eqalign{ & P = {i^2}R = {\left( {\frac{{2E}}{{R + 2r}}} \right)^2}R\,\,\,\left[ {I = \frac{E}{{{R_{{\text{eq}}}}}}} \right] \cr & = \left( {\frac{{2 \times 2}}{{2 + 2}}} \right) \times 2 = 2\,W\,\,\left[ {\because r = 1\,\Omega } \right] \cr} $$
NOTE
In case of batteries connected in series, equivalent emf is given by $${E_{{\text{eq}}}} = {E_1} + {E_2}$$