$n_{Na_2S_2O_3}=0,0235.0,1025=2,40875.10^{-3}(mol)$
$I_2+2Na_2S_2O_3\to 2NaI+Na_2S_4O_6$
$\Rightarrow n_{I_2}=1,204375.10^{-3}(mol)$
$2PbCrO_4+6KI+8H_2SO_4\to 2PbSO_4+3I_2+3K_2SO_4+Cr_2(SO_4)_3+8H_2O$
$\Rightarrow n_{Pb}=n_{PbCrO_4}=8,029.10^{-4}(mol)$
$\to m_{Pb}=207.8,029.10^{-4}=0,166g=166mg$