Using lens formula,

For Ist case : $${1 \over f} = \left( {{a^{{\mu _g}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

where, $${R_1},{R_2}$$ = radii of refracting surfaces

$f$ = focal length

$${a^{{\mu _g}}} = {{^{{\mu _g}}} \over {^{{\mu _a}}}} = {{1.5} \over 1} = 1.5$$

So, $${1 \over {24}} = (1.5 - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$ (As f = 24 cm given)

$$ \Rightarrow {1 \over {24}} = 0.5\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

$$ \Rightarrow {1 \over {12}} = \left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

$$ \Rightarrow {1 \over {{R_1}}} - {1 \over {{R_2}}} = {1 \over {12}}$$ .... (i)

For IInd case:

$${1 \over f} = \left( {w{\mu _g} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

$$ \Rightarrow {1 \over f} = \left( {{{1.5} \over {1.33}} - 1} \right)\left( {{1 \over {12}}} \right)$$ ..... from (i)

$$ \Rightarrow {1 \over f} = {{17} \over {133}} \times {1 \over {12}} \Rightarrow f = {{133 \times 12} \over {17}}$$

$$ \Rightarrow f = 93.88$$ cm

$\approx$ 94 cm

Hence, option 1 is correct.