$\begin{array}{1 1}(A)\;0.0153 \; kg\;m^{-3} \\(B)\;2.3 \;kg m^{-3} \\(C)\;7.8 \;kgm^{-3} \\(D)\;2.5\;kgm^{-3} \end{array} $

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$d= \large\frac{M}{V}$

$\qquad= \large\frac{1.32kg}{0.17m^3}$

$\qquad= 7.764 70 kgm^{-3}$

Since the factor with lowest significant figures is 0.17 and it has two significant figures, rounding the result to two significant figures, we have the density of iron ball = $7.8 kg m^{-3}$

Hence C is the correct answer.

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