An iron piece of mass 100 g is placed inside a furnace for a long...

Let the temperature of the furnace be $T$.

Mass of the iron piece, $m_i = 100 \text{ g} = 0.1 \text{ kg}$
Specific heat capacity of iron, c_i = 470 \text{ J kg}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
Total effective mass of water (water + calorimeter), $M = 240 \text{ g} + 10 \text{ g} = 250 \text{ g} = 0.25 \text{ kg}$
Specific heat capacity of water, c_w = 4200 \text{ J kg}^{-1} ^{}^{{}^{\circ}}\textC^{-1}

According to the principle of calorimetry, heat lost by the iron piece equals the heat gained by the water and the calorimeter.

$m_i c_i (T - T_f) = M c_w (T_f - T_i)$

Substituting the given values:

$0.1 \times 470 \times (T - 60) = 0.25 \times 4200 \times (60 - 20)$

$47(T - 60) = 0.25 \times 4200 \times 40$

$47(T - 60) = 42000$

T - 60 = \dfrac{42000}{47} \approx 893.6^{}^{{}^{\circ}}\text{C}

T = 893.6 + 60 = 953.6^{}^{{}^{\circ}}\text{C}

The closest option is 950^{}^{{}^{\circ}}\text{C}.