Answer
The three largest weights are $~1.02~N~$, $~0.256~N~$, and $~0.114~N$
Work Step by Step
We can write an expression for the wave speed in the string:
$v = \lambda_n~f = \frac{(2L)(f)}{n}$, for some integer $n$
The tension in the string will be equal to the weight $W$. We can find an expression for the weight:
$\sqrt{\frac{F}{\mu}} = v$
$\sqrt{\frac{W}{\mu}} = \frac{2L~f}{n}$
$\frac{W}{\mu} = (\frac{2L~f}{n})^2$
$W = (\frac{2L~f}{n})^2~(\mu)$
$W = \left(\frac{(2)(0.42~m)(110~Hz)}{n}\right)^2~(0.120\times 10^{-3}~kg/m)$
$W = \frac{1.0245~N}{n^2}$
We can find the three largest weights:
$W_1 = \frac{1.0245~N}{1^2} = 1.02~N$
$W_2 = \frac{1.0245~N}{2^2} = 0.256~N$
$W_3 = \frac{1.0245~N}{3^2} = 0.114~N$
The three largest weights are $~1.02~N~$, $~0.256~N~$, and $~0.114~N$
