Weight of body in air,
$$
m_{1}=50 g=5 \times 10^{-2} \mathrm{~kg}
$$
Weight of body in water,
$$
m_{2}=40 \mathrm{~g}=4 \times 10^{-2} \mathrm{~kg}
$$
Buoyant force on the body,
$$
\begin{aligned}
F^{\prime} &=m_{1} g-m_{2} g \\
&=g\left(m_{1}-m_{2}\right) \\
&=g\left(5 \times 10^{-2}-4 \times 10^{-2}\right) \\
F^{\prime} &=10^{-2} \mathrm{~g} \mathrm{~N}
\end{aligned}
$$
But buoyant force is given as
$$
\begin{aligned}
F^{\prime} &=V \rho g \\
10^{-2} \mathrm{~g} &=V \rho g \\
10^{-2} &=\rho \mathrm{V} \\
\Rightarrow \quad 10^{-2} &=10^{3} \mathrm{~V} \quad \text { [From Eq. (i)] } \\
\Rightarrow \quad V &=10^{-5} \mathrm{~m}^{3} \quad\left[\because \rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right] \\
\Rightarrow \quad \text { Loss of weight in liquid }=\rho_{l} V g & {\left[\because \rho_{l}=1.5 \rho\right] } \\
W^{\prime} &=1.5 \rho \times 10^{-5} \times 10 \quad \\
=& 1.5 \times 10^{3} \times 10^{-5} \times 10 \\
=& 0.15 \mathrm{~N}
\end{aligned}
$$
$\therefore$ Loss of weight (in kg) $=\frac{w^{\prime}}{g}=\frac{0.15}{10}=0.015 \mathrm{~kg}=15 \mathrm{~g}$
$\therefore$ Weight of body in water $=50-15=35 \mathrm{~g}$