Answer
$1.84\ kg$
Work Step by Step
In this case, the wood has a part of its volume in the water, and
the lead has none of its volume in the water.
Volume of water displaced = 0.900(volume of the wood).
$(V_{wood}=\displaystyle \frac{m_{wood}}{\rho_{wood}})$
The combined mass of the lead and wood are floating,
so there is no vertical acceleration $\Rightarrow$ resultant net force is zero.
Forces acting: weight $F_{g}$ (downward) and buoyant force $F_{b}$(upward),
and, their magnitudes are equal:
\begin{aligned}
F_{g}&=F_{b}\\\\
(m_{wood}+m_{lead})g&=\rho_{water}\cdot(0.900V_{wood})\cdot g\\\\
m_{wood}+m_{lead}&=\displaystyle \rho_{water}\cdot(0.900\cdot\frac{m_{wood}}{\rho_{wood}})\\\\
m_{lead}&=\displaystyle \rho_{water}\cdot(0.900\cdot\frac{m_{wood}}{\rho_{wood}})-m_{wood}\\\\
&=\displaystyle \frac{0.900\cdot(1000\ \frac{kg}{m^{3}})(3.67\ kg}{600\ \frac{kg}{m}}-3.67\ kg\\\\
&=1.84\ kg
\end{aligned}
