Price is proportional to the square of the length of wire i.e. $P\propto L^{2}$
$P_{1}=1600,L_{1}=10m,L_{2}=4m,L_{3}=6m$
We can write like this $\frac{P_{1}}{P_{2}}=\frac{L_{1}^{2}}{L_{2}^{2}}$
$\implies \frac{1600}{P_{2}}=\frac{10^{2}}{4^{2}}$
$\implies P_{2}=\frac{1600\times 16}{100}=256$
and $\frac{P_{1}}{P_{3}}=\frac{L_{1}^{2}}{L_{3}^{2}}$
$\implies\frac{1600}{P_{3}}=\frac{10^{2}}{6^{2}}$
$\implies P_{3}=\frac{1600\times 36}{100}=576$
$\implies P_{1}+P_{2}=256+576=832$
The correct answer is $(B)$
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Alternate method$:$
The price of wire made of super alloy material is proportional to the square of its length.
$\text{Price}\propto (\text{Lenght})^{2}$
$\implies P=k\ l^{2}$
Length of wire $=10 \ m$
According to the given condition,
$1600=k \ (10)^{2}\implies1600=k(100)\implies k=16$
So, that price of two wires of length $4\ m$ and $6 \ m$ is $=k\left[l_{1}^{2}+l_{2}^{2}\right]=16\left[4^{2}+6^{2}\right]=16\left[16+36\right]=16\times 52=832$
