Answer
$$1.07\times 10^3\;g$$
Work Step by Step
In this case, the can will be completely submerged in water and the lead shot will carry the can without sinking in water
Thus,
The total weight of the can and the lead shot=The buoyant force on the can
or, $(m_c+m_l)g=\rho_wVg$
or, $(m_c+m_l)=\rho_wV$
or, $m_l=\rho_wV-m_c$
Substituting the given values
$m_l=(1\times 1200-130)\;g$
or, $m_l=1070\;g$
or, $m_l=1.07\times 10^3\;g$
Therefore, the required mass of the lead is $1.07\times 10^3\;g$
