A steel sample was analysed for nickel A steel sample ($\pu{0.3500 g}$) was dissolved in nitric acid and the resulting solution made up to $\pu{500 ml}$. The absorbance of this solution measured by atomic absorption was found to be $0.410$. A standard $\pu{6 ppm}$ solution of $\ce{Ni}$ gave an absorbance of $0.522$ under the same conditions. Assuming linearity between absorbance and concentration, calculate the $\% \ce{Ni}$ in the steel sample. (8 marks)

Seems so simple but is confusing me.
A steel sample ($\pu{0.3500 g}$) was dissolved in nitric acid and the resulting solution made up to $\pu{500 ml}$.The absorbance of this solution measured by atomic absorption was found to be $0.410$.
A standard 6 ppm solution of Ni gave an absorbance of $0.522$ under the same conditions.
In the final diluted sample,
$$\frac{\pu{6 ppm}\ \ce{Ni}}{0.410\times0.522}=\pu{28.03476311 ppm}\ \ce{Ni} \tag{1 }$$
In the first solution
$$\frac{\pu{28.03476311 ppm}\ \ce{Ni}}{\pu{500 ml}}=\pu{0.05606952622 mg}\ \ce{Ni} \tag{2}$$
Supposing the density of the first solution to be near that of water:
$$\pu{1000 g}\times\pu{0.05606952622\times10^{-6} g}=\pu{5.60695\times10^{-5} g} \tag{3}$$
In the steel sample:
$$\frac{\pu{5.60695\times10^{-5} g}}{\pu{0.35 g}}=0.000160198=0.016\ \%\ \ce{Ni}\tag{4}$$
$2$nd attempt:
\begin{align}
\frac{\pu{6 ppm}}{0.522}&amp;=\frac x{0.410}\\
\implies \frac{\pu{6 ppm}\times0.410}{0.522}&amp; =x\\
x&amp;=\pu{4.7126ppm}\ \ce{Ni} \tag{6}
\end{align}
This means you have $\pu{4.7126 mg}$ nickel per $\pu{1 kg}$ of steel (ppm is one millionth, and thus, mg per kg). Then one tenth of that amount of steel $(\pu{0.1 kg}/\pu{1 kg})$ contains $1/10$ of the $\ce{Ni}$. that is $\pu{0.47126 mg}$
Since it was marked up to $\pu{500 ml}$
In the steel sample:
\begin{align}
\pu{0.47126 mg}\times\frac{500}{50}&amp;=\pu{4.7126 mg ml-1}\\
\frac{4.7126}{100}&amp;=0.047126\\
\frac{\pu{0.047126 g}}{\pu{0.35 g}}&amp;=0.134645\\
&amp;=13.46\ \%\ \ce{Ni} \tag{7}
\end{align}
I feel that I am missing a step in calculations. I would prefer to read a resource of working this out myself than finding the answer.
I feel more confident in my second attempt of finding a correct answer.
This question is from a past paper.
Your first step is incorrect. Think about it this way:
$5$ people have $10$ eyes. This group of people has $14$ eyes. How many people in the group?
The correct solution is:
This group has $14/10 = 1.4$ times more eyes than the known group. Thus, they have $1.4$ times more people. $5$ people $\cdot$($14$ eyes/$10$ eyes) $ = 5$ people $ \times 1.4$ (unitless ratio) $= 7$ people. Here, the units make sense. This is correct.
In your calculation you do ($5$ people)/($10$ eyes $\times 14$ eyes)$= 1$ people/($28$ eyes$^2$). This unit people/eyes$^2$ makes no sense. This is incorrect.
First you find the amount of $\ce{Ni}$ in the final sample.
Your sample absorbs $0.410$ which is slightly less than $0.522$ ($\pu{6 ppm }$sample which). So your sample has $(0.410/0.522) \times \pu{6 ppm} = \pu{4.713 ppm}.$
Water solution has density of $\pu{1 kg/L}$, so $\pu{500 mL} \to \pu{500 g}$. $\pu{4.713 ppm}$ in $\frac{\pu{500 g}}{1\,000\,000} = \pu{0.5 mg}$. $\pu{0.5 mg} \cdot 4.713 = \pu{2.356 mg}$.
The total weight of the sample was $\pu{0.35 g} = \pu{350 mg}$
$$100\ \% \cdot \frac{\pu{2.356 mg}}{\pu{350mg}} = 1.003\ \%$$
You should round it down to $2$ digits, because that was the accuracy of other data. The answer is  $1.00\ \%$.