Answer
$2.01\ kg$
Work Step by Step
In this case, the wood has a part of its volume in the water, and
the lead has all of its volume in the water.
Volume of water displaced = 0.900(volume of the wood)+(volume of the lead).
$(V_{wood}=\displaystyle \frac{m_{wood}}{\rho_{wood}}, \quad V_{lead}=\frac{m_{lead}}{\rho_{lead}})$
The combined mass of the lead and wood is 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{align*}
(m_{wood}+m_{lead})g&=\rho_{water}[(0.900V_{wood})+V_{lead}]\cdot g\\\\
m_{wood}+m_{lead}&= \displaystyle \rho_{water}\cdot (\frac{0.900m_{wood}}{\rho_{wood}}+\frac{m_{lead}}{\rho_{lead}})\\ \\
m_{lead}- \displaystyle \rho_{water}(\frac{m_{lead}}{\rho_{lead}})&=\displaystyle \rho_{water}\cdot (\frac{0.900m_{wood}}{\rho_{wood}})-m_{wood}\\ \\
m_{lead}(1-\displaystyle \frac{\rho_{water}}{\rho_{lead}})&=\displaystyle \rho_{water}\cdot (\frac{0.900m_{wood}}{\rho_{wood}})-m_{wood}\\ \\
m_{lead}&=[(1000\displaystyle \frac{kg}{m^{3}})\cdot (\frac{0.900(3.67kg)}{(600\frac{kg}{m^{3}})})-(3.67kg)]\div(1-\frac{(1.00\times 10^{3}\frac{kg}{m^{3}})}{(1.13\times 10^{4}\frac{kg}{m^{3}})})\\
&=2.01kg
\end{align*}