For the combination to just float, the total weight must equal the weight of displaced water. Let $m_l$ be the mass of lead. $(m_l + 10)g = (V_l + V_c) * ρ_w * g$. This simplifies to $m_l + 10 = (m_l/11) + (10/0.2)$. Solving for $m_l$ gives $m_l = 44g$.