Weight of the solid in air (W_{1}) = 44.5 gf

Density of the iron = 8.9 x 10^{3} kg m^{-3}

RD of iron = 8.9 × 1000\over 1000

= 8.9

RD = W_1\over W_1 - W_2

⇒ 8.9 = 44.5\over 44.5 - W_2

⇒ 8.9 (44.5 – W_{2}) = 44.5

⇒ 396.05 – 8.9 W_{2} = 44.5

⇒ 8.9 W_{2} = 396.05 – 44.5

⇒ W_{2}^{ }= 351.55\over 8.9 = 39.5 gm

Hence,

**Weight of iron piece when immersed in water = 39.5 gm**