Answer
The mass of the lead is $8.68~kg$
Work Step by Step
We can find the volume of the wood:
$V_w = (0.330~m)^3 = 0.03594~m^3$
We can find the mass of the wood:
$m_w = \rho_w~V_w = (780~kg/m^3)(0.03594~m^3) = 28.03~kg$
If the water just covers the lead and wood, then the average density of the lead and wood must be equal to the density of water, which is $1000~kg/m^3$.
Let the mass of the lead be $m_l$. Then the volume of the lead is $\frac{m_l}{\rho_l}$. We can find the mass of the lead $m_l$:
$\frac{m_w+m_l}{V_w+V_l} = 1000~kg/m^3$
$\frac{m_w+m_l}{V_w+\frac{m_l}{\rho_l}} = 1000~kg/m^3$
$m_w+m_l = (1000~kg/m^3)(V_w+\frac{m_l}{\rho_l})$
$m_l = (1000~kg/m^3)(V_w+\frac{m_l}{11340~kg/m^3})-m_w$
$m_l-0.0882~m_l = (1000~kg/m^3)(0.03594~m^3)-28.03~kg$
$0.9118~m_l = 35.94~kg-28.03~kg$
$m_l = \frac{7.91~kg}{0.9118}$
$m_l = 8.68~kg$
The mass of the lead is $8.68~kg$.
