What mass of lead(II) nitrate is present in 20.0mL of a solution?
See below.
Two different lead(II) nitrate solutions are used, each sample is reacted with an excess quantity of a potassium iodide solution, which produces lead(II) iodide.

This question cannot be answered because you did not specify which solution.
(Saturated? One molar?. )

You edited your question.
You should have written a new reply, because I'd not have seen it if I didn't search for it.
"two different solutions" is not a specification.
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However, lead(II)iodide is one of the least solvable salts.
The answer 0 (zero) will be very close to the truth - but not quite.
There will always be a few ions around.
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I think that your teacher wants the answer "zero".

Samples of two different lead(II) nitrate solutions are used, each sample is reacted with an excess quatity of a potassium iodide solution, producing lead(II) iodide, which has a low solubility and settle4s to the bottom of the beaker. After the contents of the beaker are filtered and dried, the mass of lead(II) iodide is determined. The reference data:
Mass of PbI2 Produced Grams --> Mass of Pb(NO3)2 Reacting

1.39 --> 1.00
2.78 --> 2.00
4.18 --> 3.00
5.57 --> 4.00
6.96 --> 5.00

This reference data relates the mass of Pb(NO3)2 to the mass of PbI2 for the reaction.

Solution 1 Solution 2
Volume Used 20.0mL 20.0mL
Mass of Filter paper 0.99g 1.02g
Mass of paper & 5.39g 8.57g
Dried precipiate

What mass of lead(II) nitrate is present in 20.0 mL of a solution?

I don't know what these figures are for, I don't need them:
1.39 --> 1.00
2.78 --> 2.00
4.18 --> 3.00
5.57 --> 4.00
6.96 --> 5.00
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Molecular weight of PbI2 is 207.7 + 2(126.9)= 461.5
Molecular weight of Pb(NO3)2 is 207.7 + 2*14 + 6*16 = 331.7
Given the equation:
Pb(NO3)2 + 2 I- => 2 NO3- + PbI2
One mole Pb(NO3)2 will produce one mole PbI2, or:
=> 331.7 gram of Pb(NO3)2 will produce 461.5 gram of PbI2
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Solution 1: found 5.39 - 0.99 = 4.4 g PbI2
Originally present: 4.4 * 331.7/461.5 = 3.16247 g Pb(NO3)2
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Solution 2: found 8.57 - 1.02 = 7.55 g PbI2
Originally present: 7.55 * 331.7/461.5 = 5.42651 g Pb(NO3)2