The solubility product of \\( \mathrm{PbI_2} \\) is \\( 8.0 \times 10^{-9} \\). The solubility of lead iodide in \\( 0.1 \\) molar solution of lead nitrate is \\( x \times 10^{-6} \, \text{mol/L} \\). The value of \\( x \\) is __________. (Rounded off to the nearest integer)
[Given \\( \sqrt{2} = 1.41 \\)]
